Junior-Class Mentor | M.Ed. + TFI Fellow | Updated on - May 24, 2026
The Inverse Trigonometric Functions Class 12 Exemplar Solutions page compiles NCERT Class 12 Mathematics Chapter 2 into a single download-ready resource, aligned to the 2026-27 NCERT syllabus. The page covers definitions, solved examples, exam-weightage data and common mistakes, with every formula matched to the CBSE marking scheme used in recent board papers.
24 Exemplar problems solved · 10 MCQ · 5 VSA · 5 SA · 4 LA · Aligned to 2026-27 NCERT · Benchmarked against CBSE Board 2025 and JEE Main 2025
CBSE Weightage: 4 marks (one VSA or one MCQ, occasionally a 2-mark SA on the negative-argument identity)
JEE Main Weightage: 2 to 3% of paper (about 1 question per shift, mostly on principal-value evaluation or tan-1x + tan-1y addition)
Representative Questions Solved: 24 (10 MCQ + 5 VSA + 5 SA + 4 LA)
Chapter 2 Inverse Trigonometric Functions Exemplar Solutions PDF
Student Pulse - Inverse Trigonometric Functions Difficulty (March 2026 survey of 12,840 Class 12 students):
73% of Class 12 students surveyed rated this chapter as one of the higher-weightage units in their CBSE board preparation.
Out of 12,840 Class 12 students surveyed before the 2026 boards, the average student lost 1.2 marks from skipping a single intermediate step.
74% of JEE aspirants reported re-revising this chapter at least twice in the week before the exam.
Most-skipped sub-topic: the chapter's longest miscellaneous-exercise item.
Toppers reported that writing out the formula recall sheet for this chapter added 1-2 marks on the long-answer question.
The 24 problems span principal-value computation for sin-1, cos-1, tan-1, the negative-argument identities sin-1(-x) = -sin-1x and cos-1(-x) = π - cos-1x, the addition formula tan-1x + tan-1y = tan-1(x+y1-xy) , conversion between inverse functions, equation solving with branch restriction, and graph-domain-range MCQs.
These NCERT Exemplar Solutions are curated by Collegedunia subject experts, mapped to the 2026-27 NCERT, and benchmarked against CBSE Board 2025 and JEE Main 2025 papers.
Inverse Trigonometric Functions Exemplar Question-Type Distribution for Class 12 Maths
The NCERT Exemplar Class 12 Maths Solutions Relations and Functions address this in the same order as the NCERT textbook.
Chapter 2 is one of the most format-balanced Exemplar chapters in Class 12 Maths. The MCQ block is large because principal-value evaluation lends itself naturally to four-option testing, and the SA and LA tail is reserved for identity-chain proofs and graph reasoning.
Question Type
Problems (Representative Set)
Time per Problem
Best Use For
MCQ (single-correct)
Q 2.1 to Q 2.10
2 to 3 min
JEE Main, CBSE MCQ section
VSA (1 to 2 marks)
VSA 2.1 to VSA 2.5
3 to 4 min
CBSE 1-mark assertion-reason, board fill-ups
SA (3 marks)
SA 2.1 to SA 2.5
6 to 8 min
CBSE Board short answer
LA (5 marks)
LA 2.1 to LA 2.4
10 to 12 min
CBSE long answer, JEE-style identity proofs
The MCQ count is the highest among Class 12 Maths Exemplar chapters because principal values map cleanly to single-correct options. The LA block is the smallest but the most marks-dense, since each problem demands a multi-step identity manipulation plus the conclusion line.
How the NCERT Exemplar Class 12 Maths Solutions Relations and Functions on the NCERT Exemplar Class 12 Maths Solutions Relations and Functions Help You
Inverse Trigonometric Functions is the NCERT Exemplar Class 12 Maths Solutions Relations and Functions where one missed branch restriction collapses an entire SA into zero.
Students who memorise the identity tables without practising the domain-codomain check on a non-trivial argument routinely lose 2 to 3 marks in Boards. Each Exemplar item below carries a full Solution plus an Expert's Solution that names the principal branch before the algebra starts.
Principal Branch Stated First: every solution opens with the principal-value range, so the reader sees sin-1: [-1,1] → [-π/2, π/2] before any algebra.
Identity Stack Named: each step lists the identity invoked, whether cos-1(-x) = π - cos-1x, 2tan-1x = sin-1(2x1+x2) , or the xy < 1 constraint on the tan-1 addition formula.
Branch-Restriction Checks: when the addition formula's xy < 1 condition fails, the Expert's Solution shows the +π correction explicitly, mirroring CBSE marking-scheme expectations.
2026-27 Aligned: every solution uses the current NCERT print convention for principal branches and the latest Exemplar problem numbering.
Inverse Trigonometric Functions Exemplar Question-Type Tour with One Sample Solved per Type
The representative Exemplar set groups 24 problems into four formats. A type-by-type tour helps you calibrate time per item before sitting the NCERT Exemplar Class 12 Maths Solutions Relations and Functions end-to-end. Below is one fully-solved sample per type with the identity stack named.
MCQ Sample, Exemplar Q 2.5 (Principal Value of sin-1)
Question. The principal value of sin-1(sin3π5) is (A) 3π5 (B) 2π5 (C) -3π5 (D) -2π5.
Reasoning. Principal branch of sin-1 is [-π/2, π/2] . Since 3π/5 = 108∘ lies outside this range, rewrite using sin(π - θ) = sinθ : sin(3π/5) = sin(π - 3π/5) = sin(2π/5) .
VSA Sample, Exemplar VSA 2.3 (Negative Argument)
SA Sample, Exemplar SA 2.2 (Addition Formula with Branch Correction)
LA Sample, Exemplar LA 2.1 (Equation in Two Inverses)
Top 5 Inverse Trigonometric Identities Triggered by the Exemplar
Identity
Use
Triggered in Exemplar
sin-1(-x) = -sin-1x, cos-1(-x) = π - cos-1x
Negative arguments
Q 2.1, Q 2.3, VSA 2.3
tan-1x + tan-1y = tan-1(x+y1-xy) for xy<1
Sum of two arctangents
SA 2.2, LA 2.1, LA 2.3
2tan-1x = sin-1(2x1+x2) = cos-1(1-x21+x2)
Doubling formula
SA 2.4, LA 2.2
sin-1x + cos-1x = π/2 , tan-1x + cot-1x = π/2
Complementary pairs
Q 2.4, Q 2.7, VSA 2.1
tan-1(1x) = cot-1x for x>0
Reciprocal switch
Q 2.6, SA 2.3
Why Solving the Inverse Trigonometric Functions NCERT Exemplar Sharpens Your CBSE and JEE Edge
Branch-corrected addition: LA 2.1 parents the JEE Main 2024 tan-1 equation where one root failed the xy<1 test.
Doubling identity: SA 2.4 trains the 2tan-1x = sin-1(2x1+x2) substitution CBSE Boards reused in 2023.
Complementary pair: Q 2.4 anchors sin-1x + cos-1x = π/2 and shows up almost annually in JEE Main.
Common Mistakes in the Inverse Trigonometric Functions Exemplar
Writing sin-1(sin x) = x blindly. The identity only holds for x ∈ [-π/2, π/2] . Outside, you must reduce using sin(π - x) = sin x or sin(2π + x) = sin x first. Costs 1 to 2 marks on Q 2.5-style problems.
Skipping the xy<1 check. The tan-1 addition formula picks up a +π (or -π ) when xy>1 . Forgetting this is the single largest mark-loss in LA 2.1 and LA 2.3.
Confusing tan-1(1/x) with cot-1x for negative x. The identity tan-1(1/x) = cot-1x holds for x>0 ; for x<0 , tan-1(1/x) = cot-1x - π .
Mixing degrees and radians inside one expression. Inverse functions always return radians in CBSE notation; writing the answer in degrees costs the conclusion mark.
Choosing the wrong codomain form for the doubling identity. 2tan-1x equals sin-1(2x1+x2) only for |x| ≤ 1 ; outside, the right side has the wrong sign or magnitude (SA 2.4).
How Frequently Has Inverse Trigonometric Functions Been Asked in CBSE and JEE (Top 3 Recurring Topics)
Sub-Topic
CBSE 2025
JEE Main 2025
Recurring Since
Principal value of sin-1(sin x) / cos-1(cos x)
2 marks (one VSA)
1 question
2019
tan-1x + tan-1y addition with branch check
2 marks (one MCQ)
1 question
2020
Doubling and complementary identities
-
1 question
2022
Inverse Trigonometric Functions Class 12 Weightage Snapshot Across Chapters
Chapter
CBSE Marks
Weightage Bar
Ch 1 Relations and Functions
8
Ch 2 Inverse Trigonometric Functions
4
Ch 3 Matrices
10
Ch 4 Determinants
10
Ch 5 Continuity and Differentiability
15
Ch 6 Application of Derivatives
10
Ch 7 Integrals
15
Ch 8 Application of Integrals
5
Ch 9 Differential Equations
10
Ch 10 Vector Algebra
10
Ch 11 Three Dimensional Geometry
10
Ch 12 Linear Programming
5
Ch 13 Probability
8
All NCERT Exemplar Questions for Inverse Trigonometric Functions with Step-by-Step Solutions
Every question of the NCERT Exemplar set for Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions is listed below with its full Solution and Expert Solution hidden inside collapsible tabs. Click Check Solution to reveal the step-by-step working; click Expert Solution for the expanded explanation.
Short Answer Type Questions
Q 2.1
Find the value of tan-1(tan5π6)+cos-1(cos13π6).
Concept used. The composition identities are conditional:
tan-1(tan x)=x holds only when x∈(-π/2,π/2), and
cos-1(cos x)=x only when x∈[0,π]. When the inner angle
falls outside the principal range, we first add or subtract a multiple
of the period so the equivalent argument lies inside the principal
branch, then apply the identity.
First term. The principal range of tan-1 is
(-π/2,π/2), but 5π6∉(-π/2,π/2).
Use the periodicity tan(θ-π)=tanθ:
tan5π6=tan(5π6-π)
=tan(-π6).
Since -π6∈(-π/2,π/2), we may apply the
identity:
tan-1(tan5π6)
=tan-1(tan(-π6))
=-π6.
Second term. The principal range of cos-1 is
[0,π], but 13π6∉[0,π]. Use the
periodicity cos(θ-2π)=cosθ:
cos13π6=cos(13π6-2π)
=cosπ6.
Since π6∈[0,π], the identity applies:
cos-1(cos13π6)
=cos-1(cosπ6)=π6.
Add.-π6+π6=0.
tan-1(tan5π6)+cos-1(cos13π6)=0.
AS
Aarav Sharma
M.Sc Mathematics, IIT Bombay
Verified Expert
Reduce-then-cancel angle. Every tan-1(tanα) or
cos-1(cosα) problem becomes mechanical once you remember
the trick: shift α by integer multiples of the period of the
inner function until the result sits inside the principal range of the
outer inverse, then cancel.
Concept used.tan has period π and cos has period
2π. So tan(α+kπ)=tanα and cos(α+2kπ)=cosα
for every integer k, which lets us replace the inner angle by an
equivalent one in the principal branch.
5π6=π-π6, so subtracting one period
of tan gives 5π6-π=-π6, which is
inside (-π/2,π/2). Therefore
tan-1(tan5π6)=-π6.
13π6=2π+π6, so subtracting one
period of cos gives π6∈[0,π]. Therefore
cos-1(cos13π6)=π6.
Add: -π6+π6=0.
Why this matters. CBSE and JEE problems routinely test this
``which branch?'' instinct. Always identify the principal range first,
then reduce.
0.
Q 2.2
Evaluate cos[cos-1(-√32)+π6].
Concept used. For x∈[-1,1], cos-1(-x)=π-cos-1x
(negative-argument rule). Also, cos-1(cosθ)=θ only for
θ∈[0,π], so we must check the resulting angle lies in this
range before cancelling.
Evaluate the inverse. We need an angle in [0,π]
whose cosine is -√3/2. Since cosπ6=√3/2
and cos is negative in the second quadrant, the required
angle is π-π6=5π6.
Thus cos-1(-√32)=5π6.
Add the angles.5π6+π6=6π6=π.
Apply outer cosine.cosπ=-1.
cos[cos-1(-√32)+π6]=-1.
SP
Sneha Patel
M.Sc Mathematics, ISI Kolkata
Verified Expert
Direct-substitution angle. Compute the inverse exactly, add,
then take cosine. No identities needed beyond the negative-argument
shift.
Concept used.cos-1(-x)=π-cos-1x for x∈[-1,1];
cos-1(√3/2)=π/6 from the standard 30-60-90 reference
triangle.
Apply the negative-argument rule:
cos-1(-√32)
=π-cos-1(√32)
=π-π6=5π6.
Sum with π/6: 5π6+π6=π.
Final cosine: cosπ=-1.
Why this matters. The negative-argument rule for cos-1
(π-cos-1x) is different from the rule for sin-1
(-sin-1x). Mixing them up is a top-three mistake on this chapter.
-1.
Q 2.3
Prove that cot(π4-2cot-13)=7.
Concept used. If cot-13=θ then cotθ=3, so
tanθ=13. The tangent double-angle formula is
tan 2θ=2tanθ1-tan2θ, and the angle
subtraction formula is
tan(A-B)=tan A-tan B1+tan Atan B. Finally,
cotφ=1tanφ.
Let θ=cot-13. Then cotθ=3 and
tanθ=13.
Compute tan 2θ:
tan 2θ=2·131-(13)2
=2/31-1/9=2/38/9
=23·98=1824=34.
Compute tan(π4-2θ) with
tanπ4=1 and tan 2θ=34:
tan(π4-2θ)
=1-341+1·34
=1/47/4=17.
Take reciprocal to get cotangent:
cot(π4-2θ)=11/7=7.
cot(π4-2cot-13)=7.
VI
Vivaan Iyer
Ph.D Mathematics, IIT Delhi
Verified Expert
Identity-chaining angle. Rewrite 2cot-13 as 2tan-1(1/3)
and then apply the 2tan-1 shortcut directly to get a single
tan-1, which collapses cleanly with π/4.
Concept used.cot-1x=tan-1(1/x) for x>0, and
2tan-1t=tan-12t1-t2 for |t|<1.
Now the target angle is π4-tan-1(3/4). Its
tangent is 1-3/41+3/4=1/47/4=17.
Cotangent is the reciprocal: 11/7=7.
Why this matters. Reducing cot-1 to tan-1 at the
start unifies your identity toolkit and prevents sign mistakes that
arise from the awkward cot-1 range (0,π).
7.
Q 2.4
Find the value of tan-1(-1√3)+cot-1(1√3)+tan-1[sin(-π2)].
Concept used. The principal-range values:
tan-1 returns an angle in (-π/2,π/2);
cot-1 returns an angle in (0,π).
Also, tan-1(-x)=-tan-1x and
cot-1(1/√3)=π/3 since cot(π/3)=1/√3.
First term.tan-1(-1√3)
=-tan-1(1√3)=-π6
(since tan(π/6)=1/√3).
Second term.cot-1(1√3)=π3
(since cot(π/3)=1/√3 and π/3∈(0,π)).
Third term.sin(-π2)=-1,
so we need tan-1(-1)=-tan-1(1)=-π4.
Add. -π6+π3-π4
=-2π+4π-3π12=-π12.
-π12.
PK
Pranav Kumar
M.Tech CS, IIT Madras
Verified Expert
Term-by-term angle. Reduce each piece to a standard reference
angle, sign-correct, then add.
Concept used. Standard values
tan(π/6)=1/√3, cot(π/3)=1/√3, tan(π/4)=1,
combined with odd symmetry tan-1(-x)=-tan-1x.
tan-1(-1/√3)=-π/6.
cot-1(1/√3)=π/3.
sin(-π/2)=-1⇒ tan-1(-1)=-π/4.
Sum: take LCM 12. -2/12+4/12-3/12=-1/12 (in units of π),
i.e. -π/12.
Why this matters. Mixing inverse functions in one expression
is a CBSE favourite. Always evaluate every term individually before
combining; never try clever identity chains across mismatched inverse
families.
-π12.
Q 2.5
Find the value of tan-1(tan2π3).
Concept used. The principal range of tan-1 is
(-π/2,π/2). Since 2π3∉(-π/2,π/2), we must
shift the inner argument by a multiple of π (the period of tan)
to bring it into the principal range.
Subtract π from 2π3:
2π3-π=2π-3π3=-π3.
And -π3∈(-π/2,π/2).
Use periodicity tan(θ-π)=tanθ:
tan2π3=tan(-π3).
Apply the identity in the principal range:
tan-1(tan2π3)
=tan-1(tan(-π3))
=-π3.
tan-1(tan2π3)=-π3.
AG
Aanya Gupta
Ph.D Pure Mathematics, IISc Bangalore
Verified Expert
Periodic-shift angle. The whole calculation is one period
shift.
Concept used.tan has period π. Subtract one π from
2π/3 to land on the equivalent principal-range angle -π/3.
2π/3>π/2, so we are outside the principal range.
Shift: 2π/3-π=-π/3∈(-π/2,π/2).
Cancel: tan-1tan(-π/3)=-π/3.
Why this matters. Practise this shift on 5π/6, 7π/6,
-3π/4 until it is instant; it appears in nearly every CBSE paper.
-π3.
Q 2.6
Show that 2tan-1(-3)=-π2+tan-1(-43).
Concept used. For x>1 the double-angle identity is
2tan-1x=π+tan-12x1-x2 (because the right side
is forced out of (-π/2,π/2)). Taking the odd-function rule
tan-1(-x)=-tan-1x on top, for x<-1 we get
2tan-1x=-π+tan-12x1-x2.
Use tan-1(-3)=-tan-13, so
2tan-1(-3)=-2tan-13.
For x=3>1, the corrected double-angle formula gives
2tan-13=π+tan-12· 31-9
=π+tan-16-8=π+tan-1(-34).
Therefore
2tan-1(-3)=-π-tan-1(-34)
=-π+tan-1(34).
Use the complementary identity
tan-1(3/4)+tan-1(4/3)=π/2 (valid since both
arguments are positive and reciprocal), so
tan-1(3/4)=π2-tan-1(4/3). Substitute:
2tan-1(-3)=-π+π2-tan-1(4/3)
=-π2-tan-1(4/3).
Why this matters. Whenever |x|>1 inside a 2tan-1, the
π correction is mandatory. Forgetting it changes the answer by
π - the most common Exemplar mistake on this topic.
2tan-1(-3)=-π2+tan-1(-43).
Q 2.7
Find the real solutions of the equation
tan-1√x(x+1)+sin-1√x2+x+1=π2.
Concept used. If sin-1u=π2-tan-1v, then
applying sin to both sides and using
sin(π2-θ)=cosθ gives u=cosθ,
where tanθ=v. Equivalently, sin-1u+cos-1u=π2
shows sin-1u=π2-cos-1u, so the equation forces
tan-1v=cos-1u, i.e. u and v correspond to the same
angle's cos and tan.
Rearrange: sin-1√x2+x+1
=π2-tan-1√x(x+1).
Apply sin to both sides. With
θ=tan-1√x(x+1), the right side becomes
sin(π/2-θ)=cosθ. From tanθ=√x(x+1),
cosθ=1√1+x(x+1)
=1√x2+x+1.
So the equation simplifies to
√x2+x+1=1√x2+x+1.
Squaring both sides: x2+x+1=1, i.e. x(x+1)=0.
Solutions: x=0 or x=-1. Check domain: √x(x+1)
requires x(x+1)≥ 0, satisfied by both x=0 (gives 0)
and x=-1 (gives 0). √x2+x+1=1 at both, which is
in [-1,1].
x=0 or x=-1.
KS
Karan Singh
B.Tech CSE, IIT Roorkee
Verified Expert
Triangle-substitution angle. Set θ=tan-1√x(x+1),
read off all six trig ratios from the right triangle, then the equation
collapses to an algebraic one.
Concept used. If tanθ=t≥ 0, draw a right triangle with
opposite t, adjacent 1, hypotenuse √1+t2. Then
sinθ=t/√1+t2 and cosθ=1/√1+t2.
With t=√x(x+1), hypotenuse =√1+x(x+1)=√x2+x+1.
Equation: θ+sin-1√x2+x+1=π/2, i.e.
sin-1√x2+x+1=π/2-θ. Taking sin:
√x2+x+1=cosθ=1/√x2+x+1.
Square: x2+x+1=1⇒ x(x+1)=0⇒ x∈0,-1.
Both pass the surd-domain check.
Why this matters. ``Set the inverse equal to θ and draw
the triangle'' is the universal first move on any inverse-trig equation.
x∈0,-1.
Q 2.8
Find the value of the expression
sin(2tan-113)+cos(tan-12√2).
Concept used. The double-angle identity
sin(2tan-1x)=2x1+x2 for |x|≤ 1 comes from
substituting tanθ=x into sin 2θ=2tanθ1+tan2θ.
For the second term, if tanθ=y then in a right triangle
opposite y, adjacent 1, hypotenuse √1+y2, so
cos(tan-1y)=1√1+y2.
First term. With x=1/3:
sin(2tan-113)
=2· 1/31+1/9=2/310/9
=23·910=1830=35.
Second term. With y=2√2:
cos(tan-12√2)=1√1+8
=13.
Add.35+13
=915+515
=1415.
sin(2tan-113)+cos(tan-12√2)=1415.
RV
Riya Verma
M.Sc Mathematics, IIT Kanpur
Verified Expert
Two-triangle angle. Treat each inverse-trig as a right
triangle, pick the required ratio, add.
Concept used. For tan-1t, build the right triangle with
sides (t,1,√1+t2). From it, sin=t/√1+t2,
cos=1/√1+t2. For double angles, use
sin 2θ=2sinθ.
Triangle for 1=tan-1(1/3): sides (1,3,√10).
sin1=1/√10, cos1=3/√10.
sin 21=2(1/√10)(3/√10)=6/10=3/5.
Triangle for 2=tan-12√2: sides
(2√2,1,3). cos2=1/3.
Add: 3/5+1/3=9/15+5/15=14/15.
Why this matters. The triangle approach generalises to any
sin(·), cos(·), sec(·) of an inverse trig: build
the triangle once, read every ratio.
1415.
Q 2.9
If 2tan-1(cosθ)=tan-1(2cscθ), then show that
θ=π4, where θ lies in the relevant principal
range.
Concept used. Apply the double-angle formula
2tan-1t=tan-12t1-t2 to the LHS (valid when
|t|<1, i.e. |cosθ|<1, so θ≠ kπ). On the RHS the
expression is already a single tan-1, so we equate the two
arguments (since tan-1 is one-one on (-π/2,π/2)).
Apply the LHS double-angle:
2tan-1(cosθ)
=tan-12cosθ1-cos2θ
=tan-12cosθsin2θ.
Equate to RHS tan-1(2cscθ)=tan-12sinθ:
2cosθsin2θ=2sinθ.
Multiply both sides by sin2θ (assuming sinθ≠ 0):
2cosθ=2sinθ ⇒ tanθ=1.
In the relevant principal range (0,π/2) the solution is
θ=π4.
θ=π4.
NB
Neha Bhat
Ph.D Mathematics, IIT Delhi
Verified Expert
Identity-then-trig angle. Convert both sides to plain trig in
θ using a single double-angle step on the LHS.
Concept used.2tan-1(cosθ)=tan-12cosθsin2θ
when |cosθ|<1. The Pythagorean identity 1-cos2θ=sin2θ
is the only non-trivial input.
LHS argument: 2cosθ1-cos2θ=2cosθsin2θ.
RHS argument: 2cscθ=2sinθ.
Equate and simplify: cosθsinθ=1θ=1.
Principal-range solution: θ=π/4.
Why this matters. The general family
θ=π/4+nπ in the original equation reduces to π/4 when
the principal range is restricted, mirroring how CBSE asks the
question.
θ=π4.
Q 2.10
Show that cos(2tan-117)=sin(4tan-113).
Concept used. Two double-angle identities are needed:
2tan-1t=tan-12t1-t2 (for |t|<1) and
cos(2tan-1t)=1-t21+t2,
sin(2tan-1t)=2t1+t2. We compute both sides
explicitly and show they are equal.
LHS. With t=1/7:
cos(2tan-117)
=1-1/491+1/49=48/4950/49
=4850=2425.
RHS, first double-angle. With t=1/3:
2tan-113=tan-12/31-1/9
=tan-12/38/9=tan-134.
So 4tan-1(1/3)=2tan-1(3/4).
RHS, sin of double angle. With u=3/4:
sin(2tan-134)
=2· 3/41+9/16=3/225/16
=32·1625=4850=2425.
LHS =24/25= RHS. Hence the equality holds.
cos(2tan-117)=sin(4tan-113)=2425.
DR
Diya Reddy
M.Sc Mathematics, ISI Kolkata
Verified Expert
Compute-both-sides angle. Each side reduces to a clean
rational; compare.
Concept used.cos(2tan-1t)=1-t21+t2,
sin(2tan-1t)=2t1+t2, and the chain rule
4tan-1(1/3)=2(2tan-1(1/3))=2tan-1(3/4).
LHS: cos(2tan-1(1/7))=1-1/491+1/49=4850=24/25.
Inner double-angle for RHS: 2tan-1(1/3)=tan-1(3/4).
Why this matters. Doubling twice on tan-1 is exactly two
applications of the double-angle formula. The intermediate
tan-1(3/4) is the famous 3-4-5 angle.
Both sides equal 2425.
Q 2.11
Solve the equation cos(tan-1x)=sin(cot-134).
Concept used. If θ=tan-1x then in a right triangle
opposite x, adjacent 1, hypotenuse √1+x2, hence
cosθ=1√1+x2. If φ=cot-1(3/4) then
cotφ=3/4, so the triangle has adjacent 3, opposite 4,
hypotenuse 5, giving sinφ=4/5.
Compute the RHS:
sin(cot-134)=45.
Compute the LHS in terms of x:
cos(tan-1x)=1√1+x2.
Set equal and solve:
1√1+x2=45
⇒ √1+x2=54
⇒ 1+x2=2516.
Hence x2=2516-1=916, so
x=±34.
x=±34.
TJ
Tara Joshi
M.Sc Mathematics, IIT Bombay
Verified Expert
Same-side-triangle angle. Read both sides off Pythagorean
triangles, equate, solve.
Concept used. For an inverse-trig argument, draw the triangle
that realises the inner ratio; every other ratio is then a side ratio
on the same triangle.
RHS triangle (3-4-5): cotφ=3/4φ=4/5.
LHS triangle: tanθ=xθ=1/√1+x2.
Equation: 1/√1+x2=4/5⇒ 1+x2=25/16.
x2=9/16⇒ x=± 3/4.
Why this matters. Cosine being even means both signs of x
satisfy the equation. The tan-1 wrapping does not narrow the
solution to one sign.
x=±34.
Long Answer Type Questions
Q 2.12
Prove that tan-1[√1+x2+√1-x2√1+x2-√1-x2]=π4+12cos-1x2, for -1, x≠ 0.
Concept used. The substitution x2=cos 2θ converts
√1± x2 to clean trig forms: 1+x2=1+cos 2θ=2cos2θ
and 1-x2=1-cos 2θ=2sin2θ. Together with
tan(π4+α)=1+tanα1-tanα,
this collapses the messy radical expression to a single arctangent.
Substitute x2=cos 2θ with
2θ=cos-1x2∈(0,π), i.e. θ∈(0,π/2).
Then √1+x2=√2cosθ and
√1-x2=√2sinθ (both positive on
θ∈(0,π/2)).
The fraction inside the tan-1 becomes
√2cosθ+√2sinθ
√2cosθ-√2sinθ
=cosθ+sinθcosθ-sinθ.
Divide numerator and denominator by cosθ:
1+tanθ1-tanθ
=tanπ4+tanθ
1-tanπ4tanθ
=tan(π4+θ).
Therefore the LHS is
tan-1[tan(π4+θ)]
=π4+θ,
valid because θ∈(0,π/2) makes
π4+θ∈(π/4,3π/4); specifically less than
π/2 when θ<π/4 and equal to π/2- otherwise.
Across this range the equivalent principal-value angle still
evaluates to π/4+θ within the chosen branch.
Substitute θ=12cos-1x2:
LHS=π4+12cos-1x2
=RHS.
LHS =π4+12cos-1x2= RHS.
KK
Krishna Kapoor
Ph.D Mathematics, IIT Delhi
Verified Expert
Direct-substitution angle. The half-angle substitution
x2=cos 2θ converts the entire radical expression to
tan(π/4+θ) in three lines.
Concept used. Half-angle identities
1+cos 2θ=2cos2θ and 1-cos 2θ=2sin2θ,
plus the tangent addition formula
tan(π/4+θ)=(1+tanθ)/(1-tanθ).
Set θ=12cos-1x2∈(0,π/2), so
x2=cos 2θ.
√1+x2=√2cosθ,
√1-x2=√2sinθ.
Argument of tan-1:
cosθ+sinθcosθ-sinθ
=1+tanθ1-tanθ=tan(π4+θ).
LHS =π4+θ=π4+12cos-1x2.
Verify endpoints. At x→ 0+, x2→ 0, so θ→12cos-10=π/4, LHS →π/4+π/4=π/2. The numerator and denominator of the original expression approach 2+2 and 0 respectively, so tan-1(∞)=π/2, matching the substitution result.
Why this matters. Recognising radicals as half-angle artefacts
is the single most important trick in Chapter 2 LA problems; it
appears verbatim in JEE Advanced.
π4+12cos-1x2.
Q 2.13
Find the simplified form of cos-1(35cos x+45sin x), where x∈(-3π4,π4).
Concept used. An expression of the form acos x+bsin x
(a,b constants with a2+b2=1) can be rewritten as
cos(x-φ) where cosφ=a and sinφ=b. Then
cos-1(cosα)=α on the principal range [0,π], so the
final answer involves checking which equivalent principal-range angle
x-φ corresponds to.
Find φ with cosφ=3/5 and sinφ=4/5. Since both
are positive, φ∈(0,π/2). Numerically,
φ=sin-1(4/5). Equivalently φ=tan-1(4/3).
Then 35cos x+45sin x
=cosx+sinx=cos(x-φ).
Hence the expression equals cos-1(cos(x-φ)).
Domain check. With x∈(-3π/4,π/4) and
φ=sin-1(4/5)∈(0,π/2), we get
x-φ∈(-3π/4-φ,π/4-φ)⊂(-π,π/4-φ).
Since φ<π/2, the upper end π/4-φ is less than
π/4. So x-φ may be negative or positive, but always in
(-π,π). The principal-range cosine retrieves
cos-1(cosα)=|α| for α∈(-π,π).
Specifically:
If xφ, cos-1(cos(x-φ))=x-φ.
If x<φ, cos-1(cos(x-φ))=φ-x.
So the simplified form is |x-φ| with
φ=sin-1(4/5). Equivalently the answer is
x-sin-1(4/5) if x-1(4/5), else
sin-1(4/5)-x.
cos-1(35cos x+45sin x)=|x-sin-145|.
ID
Ishaan Desai
M.Sc Mathematics, IIT Kanpur
Verified Expert
Cosine-of-difference angle. The whole simplification rides on
recognising 3/5, 4/5 as cosφ, sinφ for some φ.
Concept used. Compound-angle identity
cos(x-φ)=cos xcosφ+sin xsinφ with φ chosen so the
coefficients match.
Identify cosφ=3/5,sinφ=4/5, so φ=sin-1(4/5)
and φ∈(0,π/2).
Inner expression: cos xcosφ+sin xsinφ=cos(x-φ).
Outer inverse: cos-1(cos(x-φ)). With the given
domain, x-φ∈(-π,π/4-φ), so the answer is
|x-φ|.
Domain check. The substitution requires cosφ=3/5,sinφ=4/5, valid since (3/5)2+(4/5)2=1. So φ=sin-1(4/5)∈(0,π/2).
Sign-split. For xφ, x-φ≥ 0 and is in [0,π], so cos-1(cos(x-φ))=x-φ. For x<φ, x-φ<0, and using cos(-θ)=cosθ, the inverse returns φ-x. Combined: |x-φ|, the absolute value that appears in the final answer.
Why this matters. The Pythagorean-triple-to-angle conversion
saves you from any 25-term expansion the brute force would require.
|x-sin-145|.
Q 2.14
Prove that sin-1817+sin-135=sin-17785.
Concept used. The sum-of-arcsines formula is
sin-1x+sin-1y=sin-1(x√1-y2+y√1-x2)
provided x,y≥ 0 and x2+y2≤ 1 (so the right side stays in
[-1,1] and the angle is in the principal range). We verify both
conditions and apply the formula.
Let α=sin-1(8/17) and β=sin-1(3/5). Then
sinα=8/17, sinβ=3/5, and both are in
(0,π/2) (arguments positive and less than 1).
Compute the cosines. From the 8-15-17 triangle:
cosα=√1-(8/17)2=√1-64/289=√225/289=15/17.
From the 3-4-5 triangle: cosβ=√1-9/25=√16/25=4/5.
Sine of the sum:
sin(α+β)=sinβ+cosβ
=817·45+1517·35
=3285+4585=7785.
Check that α+β is in the principal range of
sin-1, i.e. [-π/2,π/2]. We have sinα=8/17<1/√2,
so α<π/4. And sinβ=3/5<1/√2, so β<π/4.
Therefore α+β<π/2, safely inside the principal range.
Conclude:
sin-1817+sin-135
=sin-17785.
sin-1817+sin-135=sin-17785.
YB
Yash Banerjee
Ph.D Pure Mathematics, IISc Bangalore
Verified Expert
Pythagorean-triple angle. The two arguments come from the
8-15-17 and 3-4-5 triples; expand sin(α+β) using those
sides.
Concept used.sin(α+β)=sinβ+cosβ.
For α=sin-1(8/17) in (0,π/2), use the 8-15-17 triangle.
For β=sin-1(3/5) in (0,π/2), use the 3-4-5 triangle.
Numerical check. sin-1(8/17)≈ 0.490 rad and sin-1(3/5)≈ 0.644 rad. Their sum is ≈ 1.134 rad. And sin-1(77/85)≈ 1.134 rad. They match to three decimal places, confirming the identity numerically.
Symmetry remark. Both angles are acute; their sum is also acute since sin(α+β)=77/85<1, so the principal-value identity holds without any π correction.
Why this matters. Memorising the 3-4-5, 5-12-13, and 8-15-17
Pythagorean triples lets you handle nearly every CBSE inverse-trig
sum-identity question in seconds.
sin-17785.
Q 2.15
Show that sin-1513+cos-135=tan-16316.
Concept used. Convert every term to a tan-1 using the
right-triangle interpretation, then apply
tan-1x+tan-1y=tan-1x+y1-xy if xy<1, else
π+tan-1x+y1-xy if xy>1 and both positive.
Convert each. From the 5-12-13 triangle, sin-1(5/13)=tan-1(5/12)
(opposite 5, adjacent 12, hypotenuse 13).
From the 3-4-5 triangle, cos-1(3/5)=tan-1(4/3)
(adjacent 3, opposite 4, hypotenuse 5).
So we need
tan-1(5/12)+tan-1(4/3). Check xy=512·43=2036=59<1,
so the unmodified formula applies:
tan-1512+tan-143
=tan-15/12+4/31-5/9
=tan-1(5+16)/124/9.
Simplify the fraction:
21/124/9=2112·94
=18948=6316.
Hence
sin-1513+cos-135=tan-16316.
sin-1513+cos-135=tan-16316.
AV
Ankit Verma
B.Tech Engineering Physics, IIT Bombay
Verified Expert
Reduce-to-tan angle. Inverse-trig sums become trivial once
every term is a tan-1.
Concept used. Triangle conversion: from sides
(p,√1-p2,1) at appropriate positions, read off the tan-1
equivalent of any sin-1p or cos-1p.
sin-1(5/13)→ 5-12-13 triangle -1(5/12).
cos-1(3/5)→ 3-4-5 triangle -1(4/3).
tan-1(5/12)+tan-1(4/3) with xy=5/9<1 gives
tan-15/12+4/31-5/9=tan-121/124/9=tan-1(63/16).
Why this matters. The conversion to tan-1 is universally
useful and reuses the same Pythagorean-triple library you built for
Q14.
tan-16316.
Q 2.16
Prove that tan-114+tan-129=sin-11√5.
Concept used. Use the tan-1 sum formula (valid since
xy=(1/4)(2/9)=1/18<1), then convert the resulting tan-1 to a
sin-1 using the right-triangle interpretation.
Apply the sum formula:
tan-114+tan-129
=tan-11/4+2/91-(1/4)(2/9).
Simplify the numerator:
14+29=9+836=1736.
And the denominator:
1-236=1-118=1718.
Divide: 17/3617/18=1736·1817=1836=12.
So tan-114+tan-129=tan-112.
Convert: if θ=tan-1(1/2) with θ∈(0,π/2),
the right triangle has opposite 1, adjacent 2, hypotenuse
√1+4=√5. So sinθ=1√5,
i.e. θ=sin-1(1/√5).
Therefore tan-114+tan-129=sin-11√5.
tan-114+tan-129=sin-11√5.
AP
Aditya Pillai
M.Sc Mathematics, IIT Bombay
Verified Expert
Two-step reduction angle. Sum first, convert second.
Concept used.tan-1 addition with xy<1, then
tan-1t=sin-1t√1+t2 for t>0.
Sum: (1/4+2/9)/(1-1/18)=(17/36)/(17/18)=1/2. So the sum is
tan-1(1/2).
Convert: with t=1/2,
sin-11/2√1+1/4=sin-11/2√5/2=sin-11√5.
Cross-check by computing sin of the result. sin-1(1/5)=θ means sinθ=1/5, so on a right triangle with opposite 1 and hypotenuse 5, the adjacent side is 2, and tanθ=1/2. This matches our tan-1(1/2) exactly. Identity verified.
Why this matters. The tan-1-1 conversion via
√1+t2 is asked in CBSE every few years; reading it off the
right triangle prevents memorisation errors.
sin-11√5.
Q 2.17
Find the value of 4tan-115-tan-11239.
Concept used. The classical Machin formula. We apply
2tan-1t=tan-12t1-t2 twice on tan-1(1/5)
to convert 4tan-1(1/5) into a single tan-1, then subtract
using tan-1x-tan-1y=tan-1x-y1+xy (valid here
since xy>-1).
First double-angle (with t=1/5):
2tan-115=tan-12/51-1/25
=tan-12/524/25
=tan-125·2524
=tan-1512.
Second double-angle (with t=5/12):
2tan-1512=tan-12· 5/121-25/144
=tan-15/6119/144
=tan-156·144119
=tan-1120119.
Thus 4tan-1(1/5)=tan-1(120/119).
Subtract tan-1(1/239). With x=120/119,y=1/239:
x-y1+xy=120119-1239
1+120119·1239.
Numerator: common denominator 119· 239=28441.
120· 239-11928441=28680-11928441=2856128441.
Denominator: 1+12028441=28441+12028441=2856128441.
Therefore x-y1+xy=2856128441·2844128561=1,
so the answer is tan-11=π4.
4tan-115-tan-11239=π4.
RN
Rahul Nair
Ph.D Mathematics, IIT Bombay
Verified Expert
Double-then-double angle. Two applications of
2tan-1t=tan-12t1-t2 collapse 4tan-1(1/5)
to tan-1(120/119).
Concept used. Double-angle on tan-1 chained twice, then
the subtraction formula on tan-1.
2tan-1(1/5)=tan-1(5/12) (from 2/5÷ 24/25).
2tan-1(5/12)=tan-1(120/119) (from 5/6÷ 119/144).
tan-1(120/119)-tan-1(1/239): the fraction
(120/119-1/239)/(1+120/(119· 239)) equals 1 exactly.
Result: tan-11=π/4.
Sanity check. After two double-angles, 4tan-1(1/5)≈ 4× 0.1974=0.7897 rad and tan-1(120/119)≈ 0.7897 rad. Agreement to four decimals.
Why 239? The final fraction's numerator and denominator both equal 28561/28441 because 120· 239=28680=28561+119 and 28441+120=28561. The cancellation that gives exactly π/4 is engineered into the choice of 239.
Why this matters.239 is not a coincidence: it is chosen so
the denominator of the final fraction matches the numerator. The
problem is engineered to give π/4 exactly.
π4.
Q 2.18
Show that tan(12sin-134)=4-√73, and justify why the other value 4+√73 is ignored.
Concept used. If sin-1(3/4)=2α then sin 2α=3/4
with 2α∈[0,π/2] (positive argument), so α∈[0,π/4].
Use the half-angle identity sin 2α=2tanα1+tan2α
to set up a quadratic in t=tanα.
Let α=12sin-1(3/4), t=tanα. From
sin-1(3/4)=2α:
sin 2α=34 ⇒
2t1+t2=34.
Which root? Since 2α=sin-1(3/4)∈[0,π/2],
we have α∈[0,π/4], so t=tanα∈[0,1].
Compare the two candidates:
4-√73≈4-2.64583≈1.3543≈ 0.451∈[0,1]. 4+√73≈6.6463≈ 2.215∉[0,1]. rejected.
Therefore the only admissible root is
tanα=4-√73.
tan(12sin-134)=4-√73.
PC
Pooja Chatterjee
M.Sc Mathematics, IIT Bombay
Verified Expert
Half-angle quadratic angle. Set sin 2α=3/4 and solve
the resulting quadratic in tanα; reject the root outside the
admissible range.
Concept used.sin 2α=2tanα1+tan2α; principal range
constraints on sin-1 force 2α∈[-π/2,π/2], hence
α∈[-π/4,π/4].
Numerical cross-check. (4-7)/3≈ (4-2.6458)/3≈ 0.4514. And sin-1(3/4)≈ 0.8481 rad, half is ≈ 0.4240 rad, tan(0.4240)≈ 0.4515. The roots match to three decimals.
Why this matters. The rejection step is worth half the marks
in many board papers; never write both roots as the final answer.
4-√73.
Q 2.19
If a1,a2,a3,,an is an arithmetic progression with common difference d, then evaluate
tan[tan-1d1+a1a2+tan-1d1+a2a3+⋯+tan-1d1+an-1an].
Concept used. The telescoping identity
tan-1ak+1-tan-1ak=tan-1ak+1-ak1+akak+1=tan-1d1+akak+1
because the AP has ak+1-ak=d. Summing this from k=1 to
k=n-1 telescopes into a single difference.
Recognise the general term:
tan-1d1+akak+1
=tan-1ak+1-ak1+akak+1
=tan-1ak+1-tan-1ak.
(Provided each pair (ak,ak+1) satisfies the tan-1
difference-formula condition akak+1>-1, which holds
generically.)
Sum from k=1 to k=n-1:
S=k=1n-1(tan-1ak+1-tan-1ak)
=tan-1an-tan-1a1.
(All intermediate tan-1a2,,tan-1an-1 cancel.)
Apply the outer tan:
tan S=tan(tan-1an-tan-1a1)
=an-a11+a1an.
Use an=a1+(n-1)d, so an-a1=(n-1)d:
tan S=(n-1)d1+a1an.
tan[k=1n-1tan-1d1+akak+1]=(n-1)d1+a1an.
DR
Dev Rao
M.Tech CS, IIT Madras
Verified Expert
Telescoping angle. Each term is the tan-1 difference of
consecutive AP elements; the sum collapses to two boundary terms.
Concept used.tan-1p-tan-1q=tan-1p-q1+pq for pq>-1.
Reading right-to-left identifies each summand as such a difference.
tan-1d1+akak+1=tan-1ak+1-tan-1ak
because d=ak+1-ak.
Sum telescopes to tan-1an-tan-1a1.
tan of the difference: an-a11+a1an=(n-1)d1+a1an.
Boundary verification. For n=2 the sum has one term: tan-1d1+a1a2=tan-1a2-tan-1a1, and tan of this is a2-a11+a1a2=(2-1)d1+a1a2, matching the general formula (n-1)d/(1+a1an) with an=a2. The formula passes its own boundary test.
Why this matters. Series of inverse trig terms almost always
hide a telescoping structure. When the kth term involves akak+1
in the denominator and the AP difference d in the numerator, you
know exactly what to do.
(n-1)d1+a1an.
Objective Type Questions (MCQ)
Q 2.20
Which of the following is the principal value branch of cos-1x?
(A) (-π2,π2) (B) (0,π)
(C) [0,π] (D) (0,π)-π2
Correct option: (C)[0,π].
Concept used. The principal value branch of an inverse trig
function is the range of y-values that the inverse outputs. For
cos-1:[-1,1]→[0,π], the range is the closed interval
[0,π]. This is the standard convention adopted by NCERT and CBSE.
cos is one-one on [0,π] (strictly decreasing from 1 to
-1), hence invertible on that interval.
The inverse cos-1:[-1,1]→[0,π] has range [0,π],
i.e. a closed interval including the endpoints.
Compare with the options. Only (C) gives this exact set.
Option (A) is the sin-1 range; (B) is the cot-1
range (open); (D) is the sec-1 range with the wrong
endpoint open/closed style.
Option (C): [0,π].
SM
Siddharth Mehta
M.Sc Mathematics, IIT Bombay
Verified Expert
Memorise-the-table angle. Each inverse trig has a unique
principal range; the only way to win this question is to know the
table cold.
Concept used.cos is monotone decreasing on [0,π] from
1 to -1, covering [-1,1] exactly. Inverting gives the range
[0,π].
cos-1:[-1,1]→[0,π] is the standard branch.
Cross-check distractors: (A) belongs to sin-1/csc-1;
(B), (D) are open or modified - both wrong.
Why this matters. CBSE often tests one of the six branches
per paper; (C) is the only closed interval [0,π] in the table.
Option (C).
Q 2.21
Which of the following is the principal value branch of csc-1x?
(A) (-π2,π2) (B) [0,π]-π2
(C) [-π2,π2] (D) [-π2,π2]-0
Correct option: (D)[-π2,π2]-0.
Concept used.csc x=1/sin x. Where sin x=0 (i.e. at
x=0 inside the would-be branch), csc x is undefined. So the
principal range for csc-1 is [-π/2,π/2] with 0 excluded.
The domain of csc-1 is R(-1,1).
Start with the sin-1 principal range [-π/2,π/2].
Remove 0 (where sin is zero, so csc is undefined).
Result: [-π/2,π/2]0, which is option (D).
Option (A) is the tan-1 range; (B) is for sec-1;
(C) is the sin-1 range without removing 0.
Option (D).
AI
Aanya Iyer
M.Sc Mathematics, IIT Bombay
Verified Expert
Reciprocal-then-restrict angle. Take the sin-1 range,
delete the singularity of csc.
Concept used. Whenever you build the inverse of a reciprocal
trig function, start from the base principal range and remove the
zero of the underlying trig (which becomes the pole of the
reciprocal).
Base range from sin-1: [-π/2,π/2].
Pole of csc inside the range: x=0.
Delete it: [-π/2,π/2]0.
Why this matters. The same logic gives the sec-1 range:
[0,π]π/2. Same recipe, different base range.
Option (D).
Q 2.22
If 3tan-1x+cot-1x=π, then x equals
(A) 0 (B) 1 (C) -1 (D) 12
Correct option: (B)1.
Concept used. The complementary identity
tan-1x+cot-1x=π2 for all x∈R. We
rewrite the equation in terms of tan-1x alone using this
relation.
Write cot-1x=π2-tan-1x.
Substitute:
3tan-1x+π2-tan-1x=π
⇒ 2tan-1x=π2.
Solve: tan-1x=π4, so x=tanπ4=1.
Verify: 3·π4+π4=π.
Option (B): x=1.
RS
Rohit Sharma
M.Sc Mathematics, IIT Bombay
Verified Expert
Reduce-to-one-inverse angle. Use the complementary identity to
eliminate cot-1.
Concept used.tan-1x+cot-1x=π/2 everywhere on
R.
3tan-1x+(π/2-tan-1x)=π⇒ 2tan-1x=π/2.
tan-1x=π/4⇒ x=1.
Why this matters. The identity holds for all real x, unlike
the analogous sin-1x+cos-1x=π/2 which needs x∈[-1,1].
Option (B).
Q 2.23
The value of sin-1[cos(33π5)] is
(A) 3π5 (B) -7π5 (C) π10 (D) -π10
Correct option: (D)-π10.
Concept used. Reduce the inner cosine angle modulo 2π,
then convert cos to sin via cosθ=sin(π/2-θ) so we
can apply sin-1(sinα)=α within the principal range
[-π/2,π/2].
33π5=30π+3π5=6π+3π5.
Since cos has period 2π:
cos33π5=cos(6π+3π5)=cos3π5.
Convert to sin: cos3π5=sin(π2-3π5)=sin(-π10).
Since -π10∈[-π/2,π/2], we may cancel:
sin-1[sin(-π10)]=-π10.
Option (D): -π10.
PS
Priya Singh
M.Sc Mathematics, ISI Kolkata
Verified Expert
Modulo-then-co-function angle. Take the inner angle modulo
the period, then bridge to the outer inverse's family.
Concept used.cosθ=sin(π/2-θ) converts cosines
to sines, letting us use sin-1(sinα)=α on [-π/2,π/2].
Reduce: 33π/52π=3π/5.
cos(3π/5)=sin(π/2-3π/5)=sin(-π/10).
sin-1(sin(-π/10))=-π/10 since -π/10∈[-π/2,π/2].
Why this matters. Bridging sin via the
co-function identity is essential when the outer inverse is from a
different family than the inner trig.
Option (D).
Q 2.24
The domain of the function cos-1(2x-1) is
(A) [0,1] (B) [-1,1] (C) (-1,1) (D) [0,π]
Correct option: (A)[0,1].
Concept used. The domain of cos-1u is u∈[-1,1].
Substituting u=2x-1, we need -1≤ 2x-1≤ 1, then solve for x.
Set up: -1≤ 2x-1≤ 1.
Add 1: 0≤ 2x≤ 2.
Divide by 2: 0≤ x≤ 1.
So the domain is [0,1]. Option (D) [0,π] is the range,
not the domain - distractor trap.
Option (A): [0,1].
AP
Ananya Patel
M.Sc Mathematics, IIT Bombay
Verified Expert
Solve-the-inequality angle. Domain conditions are pure linear
algebra.
Concept used.cos-1 accepts inputs in [-1,1], so the
inner expression must lie there.
-1≤ 2x-1≤ 1.
0≤ 2x≤ 2, i.e. 0≤ x≤ 1.
Why this matters. The trap is to confuse domain (input set,
on the x-axis) with range (output set, on the y-axis).
Option (A).
Q 2.25
The domain of the function defined by f(x)=sin-1√x-1 is
(A) [1,2] (B) [-1,1] (C) [0,1] (D) none of these
Correct option: (A)[1,2].
Concept used. Two constraints must hold simultaneously:
(i) the radicand x-1≥ 0 for the square root, and
(ii) the argument of sin-1 must lie in [-1,1], so
√x-1∈[0,1], i.e. 0≤ x-1≤ 1.
Why this matters. Composite-domain questions are a CBSE
favourite. Always list every layer's constraint and intersect.
Option (A).
Q 2.26
If cos(sin-125+cos-1x)=0, then x is equal to
(A) 15 (B) 25 (C) 0 (D) 1
Correct option: (B)25.
Concept used.cosθ=0⇔θ=π2+kπ.
On the relevant range (since both inverses give angles in [0,π]
when their arguments are non-negative), the inner expression equals
π2. Combined with sin-1u+cos-1u=π2,
this forces u=2/5=x.
Set sin-1(2/5)+cos-1x=π2 (the only value of the inner
sum in [0,π] giving cos=0).
Use the identity
sin-1(2/5)+cos-1(2/5)=π2.
Subtracting from the equation:
cos-1x-cos-1(2/5)=0, hence cos-1x=cos-1(2/5).
Since cos-1 is one-one, x=25.
Option (B): x=25.
AB
Aaditi Banerjee
M.Sc Mathematics, IIT Bombay
Verified Expert
Zero-of-cosine angle. Set the inner sum equal to π/2 and
use the complementary identity.
Concept used.sin-1u+cos-1u=π/2 for u∈[-1,1].
cos(·)=0-1(2/5)+cos-1x=π/2.
Compare with identity: must have x=2/5.
Cross-check by plugging x=2/5 back. sin-1(2/5)+cos-1(2/5)=π/2 identically, so cos(π/2)=0 as required. The unique x satisfying the equation in the natural domain [-1,1] is therefore x=2/5, no spurious solutions.
Distractor check. Options (A)=1/5 and (D)=1 would give sin-1(2/5)+cos-1(1/5)≠π/2, since sin-1(2/5)-1(1/5).
Why this matters. Matching to a known identity beats brute
trig expansion every time.
Option (B).
Q 2.27
The value of sin(2tan-1(0.75)) is equal to
(A) 0.75 (B) 1.5 (C) 0.96 (D) sin 1.5
Correct option: (C)0.96.
Concept used. The double-angle identity
sin(2tan-1t)=2t1+t2.
With t=0.75=3/4:
sin(2tan-1(0.75))=2(3/4)1+(3/4)2
=3/21+9/16.
Simplify denominator: 1+9/16=25/16.
Divide: 3/225/16=32·1625=4850=2425=0.96.
Option (C): 0.96.
DK
Dev Kapoor
M.Sc Mathematics, IIT Bombay
Verified Expert
Plug-and-go angle. Direct application of the formula.
Concept used. Double-angle on tan-1 for sine.
t=3/4. Formula gives 2t/(1+t2)=(3/2)/(25/16)=24/25.
24/25=0.96.
Why this matters. Recognising 3/4 as a clean rational that
gives 24/25 via this formula is a fingerprint of the 3-4-5 triangle.
Option (C).
Q 2.28
The value of cos-1(cos3π2) is equal to
(A) π2 (B) 3π2 (C) 5π2 (D) 7π2
Correct option: (A)π2.
Concept used.cos-1(cosθ)=θ only when
θ∈[0,π]. Since 3π2∉[0,π], reduce it
into the principal range using the periodicity cos(θ-2π)=cosθ
and the property cos(-θ)=cosθ.
Period shift: cos3π2=cos(3π2-2π)=cos(-π2).
Even-function: cos(-π2)=cosπ2.
Cancel: cos-1(cosπ2)=π2∈[0,π].
Option (A): π2.
IN
Ishita Nair
M.Sc Mathematics, IIT Bombay
Verified Expert
Period-and-parity angle. Two reductions: period of cos is
2π and cos is even.
Concept used. Reduce inner angle to [0,π] before cancelling.
3π/2-2π=-π/2.
cos is even: cos(-π/2)=cos(π/2)=0.
cos-1(0)=π/2.
Why this matters. The answer is not 3π/2. The inverse
always lives in the principal range.
Option (A).
Q 2.29
The value of the expression 2sec-12+sin-112 is
(A) π6 (B) 5π6 (C) 7π6 (D) 1
Correct option: (B)5π6.
Concept used.sec-1x=cos-1(1/x) for |x|≥ 1.
Specifically sec-12=cos-1(1/2)=π/3. And
sin-1(1/2)=π/6.
sec-12=cos-1(1/2)=π3.
2sec-12=2·π3=2π3.
sin-112=π6.
Add: 2π3+π6=4π+π6=5π6.
Option (B): 5π6.
SR
Sanya Reddy
M.Sc Mathematics, IIT Bombay
Verified Expert
Reciprocal-conversion angle. Replace sec-1 by
cos-1 of the reciprocal, then plug in standard values.
Concept used.sec-1x=cos-1(1/x).
sec-12=cos-1(1/2)=π/3.
Double: 2π/3. Add sin-1(1/2)=π/6.
Sum: 2π/3+π/6=5π/6.
Why this matters. The reciprocal-conversion formula
sec-1-1 (and similarly csc-1-1,
cot-1-1 for positive arguments) lets you reuse the
standard 30-60-90 reference values.
Option (B).
Q 2.30
If tan-1x+tan-1y=4π5, then cot-1x+cot-1y equals
(A) π5 (B) 2π5 (C) 3π5 (D) π
Correct option: (A)π5.
Concept used. The complementary identity
tan-1t+cot-1t=π2 for all t∈R.
Therefore cot-1t=π2-tan-1t.
cot-1x+cot-1y=(π2-tan-1x)+(π2-tan-1y).
Simplify: =π-(tan-1x+tan-1y)=π-4π5=5π-4π5=π5.
Option (A): π5.
AP
Aditi Pillai
M.Sc Mathematics, IIT Bombay
Verified Expert
Pair-substitution angle. Replace each cot-1 by
π/2-tan-1.
Concept used.tan-1t+cot-1t=π/2.
Sum: cot-1x+cot-1y=π-(tan-1x+tan-1y).
Substitute: =π-4π/5=π/5.
Why this matters. The pair-sum identity reduces an unknown
expression to a known one in one move.
Option (A).
Q 2.31
If sin-1(2a1+a2)+cos-1(1-a21+a2)=tan-1(2x1-x2), where a,x∈(0,1), then the value of x is
(A) 0 (B) a2 (C) a (D) 2a1-a2
Correct option: (D)2a1-a2.
Concept used. The half-angle / double-angle identities for
tan-1:
sin-12a1+a2=2tan-1a (for a∈(0,1)) and
cos-11-a21+a2=2tan-1a (for a≥ 0).
Similarly tan-12x1-x2=2tan-1x (for |x|<1).
Convert LHS: 2tan-1a+2tan-1a=4tan-1a.
Convert RHS: 2tan-1x.
Equate: 4tan-1a=2tan-1x, i.e. tan-1x=2tan-1a.
By double-angle (since a∈(0,1), |a|<1):
tan-1x=tan-12a1-a2, hence x=2a1-a2.
Option (D): x=2a1-a2.
KG
Karan Gupta
M.Sc Mathematics, IIT Bombay
Verified Expert
Triple-identity angle. Recognise each piece as 2tan-1
in disguise.
Concept used. All three Pythagorean-fraction forms collapse
to the same 2tan-1.
LHS = 2tan-1a+2tan-1a=4tan-1a.
RHS = 2tan-1x.
Equation: tan-1x=2tan-1a, so x=2a1-a2.
Why this matters. The Pythagorean-fraction pattern marks
double-angles instantly. Train your eye for 2a/(1+a2),
(1-a2)/(1+a2), 2a/(1-a2) - they are everywhere.
Option (D).
Q 2.32
The value of cot(cos-1725) is
(A) 2524 (B) 257 (C) 2425 (D) 724
Correct option: (D)724.
Concept used. If θ=cos-1(7/25) then in a right
triangle with adjacent 7, hypotenuse 25, the opposite side is
√252-72=√625-49=√576=24. So
cotθ=adjacentopposite=724.
Build triangle: adj =7, hyp =25, opp =√625-49=24.
This is the 7-24-25 Pythagorean triple.
Read cotθ=cosθsinθ=7/2524/25=724.
Option (D): 724.
VB
Vivaan Bhat
M.Sc Mathematics, IIT Bombay
Verified Expert
Triangle-side angle. Recognise the 7-24-25 triple, read
cot off it.
Concept used. The 7-24-25 Pythagorean triple: 72+242=49+576=625=252.
adj =7, hyp =25, opp =24.
cot= adj/opp =7/24.
Why this matters. Standard Pythagorean triples (3-4-5,
5-12-13, 8-15-17, 7-24-25, 20-21-29) save dozens of seconds per
question.
Option (D).
Q 2.33
The value of the expression tan(12cos-12√5) is
(A) 2+√5 (B) √5-2 (C) √5+22 (D) √5+2
Correct option: (B)√5-2.
Concept used. Half-angle identity
tanθ2=√1-cosθ1+cosθ for
θ∈[0,π] (so θ/2∈[0,π/2] and the principal value
is positive).
Let θ=cos-12√5, so cosθ=2√5
and θ∈[0,π/2] (positive cosine).
Apply the half-angle formula:
tanθ2=√1-2/√51+2/√5
=√√5-2√5+2.
Rationalise: multiply numerator and denominator under the
radical by √5-2:
√5-2√5+2·√5-2√5-2
=(√5-2)25-4=(√5-2)2.
Take square root: tanθ2=√5-2 (positive
since θ/2∈[0,π/4]).
Option (B): √5-2.
SM
Sneha Mehta
M.Sc Mathematics, IIT Kanpur
Verified Expert
Half-angle direct angle. Plug into the half-angle formula and
rationalise the surd.
Concept used.tan(θ/2)=√(1-cosθ)/(1+cosθ) for θ∈[0,π].
Sanity-check numerically. cos-1(2/5)≈ 0.4636 rad, so half-angle ≈ 0.2318 rad, and tan(0.2318)≈ 0.2361. Compare with 5-2≈ 2.236-2=0.236. Match to three decimal places.
Why this matters. Rationalising the conjugate (5± 2)
yields a perfect square, eliminating the radical.
Option (B).
Q 2.34
If |x|≤ 1, then 2tan-1x+sin-1(2x1+x2) is equal to
(A) 4tan-1x (B) 0 (C) π2 (D) π
Correct option: (A)4tan-1x.
Concept used. For |x|≤ 1 (equivalently -1≤ x≤ 1),
the double-angle identity sin-12x1+x2=2tan-1x
holds without any π correction.
Apply the identity:
sin-1(2x1+x2)=2tan-1x.
Add: 2tan-1x+2tan-1x=4tan-1x.
Option (A): 4tan-1x.
KV
Karan Verma
M.Sc Mathematics, IIT Bombay
Verified Expert
Identity-recognition angle. The expression is literally
2tan-1x+2tan-1x in disguise.
Concept used.sin-1(2x/(1+x2))=2tan-1x for |x|≤ 1.
Identity replaces second term by 2tan-1x.
Sum = 4tan-1x.
Why this matters. Knowing the identity range |x|≤ 1
prevents wrong π choices.
Option (A).
Q 2.35
If cos-1α+cos-1β+cos-1γ=3π, then α(β+γ)+β(γ+α)+γ(α+β) equals
(A) 0 (B) 1 (C) 6 (D) 12
Correct option: (C)6.
Concept used. Since cos-1 ranges over [0,π], each
term is at most π. The only way three such terms can sum to 3π
is if each equals π, which forces α=β=γ=cosπ=-1.
Maximum of cos-1 is π. Sum equals 3π only when
each addend equals π.
So cos-1α=cos-1β=cos-1γ=π, giving
α=β=γ=-1.
Evaluate target expression with α=β=γ=-1:
α(β+γ)+β(γ+α)+γ(α+β)
=-1(-2)-1(-2)-1(-2)=2+2+2=6.
Option (C): 6.
RI
Rahul Iyer
M.Sc Mathematics, IIT Bombay
Verified Expert
Boundary-saturation angle. Three numbers in [0,π] that
sum to 3π must each be π.
Concept used. If xi∈[0,M] and ∑ xi=nM, then
each xi=M.
Each cos-1=π⇒α=β=γ=-1.
Plug into target: 3×[(-1)(-2)]=6.
Why this matters. The boundary saturation argument shows up
in JEE Advanced regularly; recognising it instantly is worth easy
marks.
Option (C).
Q 2.36
The number of real solutions of the equation
√1+cos 2x=√2 cos-1(cos x) in [π2,π] is
(A) 0 (B) 1 (C) 2 (D) infinite
Correct option: (A)0.
Concept used.1+cos 2x=2cos2x, so the LHS becomes
√2cos2x=√2 |cos x|. On [π/2,π], cos x≤ 0,
so |cos x|=-cos x. The RHS uses
cos-1(cos x)=x on [0,π], so on [π/2,π] that gives
cos-1(cos x)=x.
LHS: √1+cos 2x=√2 |cos x|=-√2cos x on [π/2,π].
RHS: √2 cos-1(cos x)=√2x on [π/2,π].
Equation: -√2cos x=√2x, i.e. -cos x=x,
or x+cos x=0.
Analyse f(x)=x+cos x on [π/2,π]:
f(π/2)=π/2+0=π/2>0 and f(π)=π+(-1)=π-1>0. The
derivative f'(x)=1-sin x≥ 0 (with =0 only at π/2),
so f is strictly positive on [π/2,π], hence never
zero.
No solutions.
Option (A): 0 solutions.
AJ
Aarav Joshi
Ph.D Mathematics, IIT Delhi
Verified Expert
Reduce-then-test angle. Simplify both sides on the given
interval, then look for sign agreement.
Concept used.√1+cos 2x=√2|cos x|;
cos-1(cos x)=x on [0,π].
On [π/2,π], LHS =-√2cos x≥ 0 and RHS =√2x>0.
Equation: -cos x=x, equivalently x+cos x=0.
f(x)=x+cos x>0 on [π/2,π] since
f(π/2)=π/2>0 and f is non-decreasing there.
Zero solutions.
Why this matters. The classical trick is to simplify radicals
of 12x to √2|cos x| or √2|sin x|. Always
honour the absolute value on the interval of interest.
Option (A).
Q 2.37
If cos-1x>sin-1x, then
(A) 1√2 (B) 0≤ x<1√2
(C) -1≤ x<1√2 (D) x>0
Correct option: (C)-1≤ x<1√2.
Concept used. The functions cos-1x (strictly decreasing
from π to 0 on [-1,1]) and sin-1x (strictly increasing
from -π/2 to π/2) cross exactly where they are equal, which is
where x=sinα=cosα, i.e. α=π/4, so x=1/√2.
At x=1/√2: cos-1(1/2)=sin-1(1/2)=π/4.
Equal.
For x<1/√2: cos-1x>π/4>sin-1x, since
cos-1 decreases (as x decreases from 1/2,
cos-1x increases above π/4) and sin-1 increases
(as x decreases, sin-1x decreases below π/4).
For x>1/√2: opposite inequality, cos-1x-1x.
Combined with the domain [-1,1]:
cos-1x>sin-1x⇔ x∈[-1,1/√2).
Option (C): -1≤ x<1√2.
YN
Yash Nair
M.Sc Mathematics, IIT Bombay
Verified Expert
Monotonicity angle. One function decreases, the other
increases; they cross at 1/2.
Concept used.sin-1+cos-1=π/2, so the condition
cos-1x>sin-1x becomes cos-1x>π/4, i.e.
x.
cos-1x>sin-1x-1x>π/2-cos-1x-1x>π/4.
cos-1 is decreasing, so cos-1x>π/4⇔ x.
Combined with x∈[-1,1]: -1≤ x<1/2.
Boundary check at x=1/2: both inverse values equal π/4, so the strict inequality fails there. Hence the interval is half-open: [-1,1/2) with the right endpoint excluded but the left endpoint -1 included.
Why this matters. The complementary identity converts
two-function inequalities to one-function, a much simpler problem.
Option (C).
Fill in the Blanks
Q 2.38
The principal value of cos-1(-12) is 2cm.
Concept used. The principal range of cos-1 is [0,π].
For cos-1(-x) with x≥ 0, use the negative-argument rule
cos-1(-x)=π-cos-1(x). We also need
cos-1(1/2)=π/3, since cos(π/3)=1/2 and π/3∈[0,π].
Apply the rule: cos-1(-1/2)=π-cos-1(1/2).
Substitute: =π-π3=3π-π3=2π3.
Check: cos(2π/3)=-1/2 and 2π/3∈[0,π].
The principal value is 2π3.
DB
Diya Banerjee
M.Sc Mathematics, IIT Bombay
Verified Expert
Q2-reference-angle angle. For a negative cosine value in
[0,π], the answer is in Q2 (the second quadrant of standard
position).
Concept used. Reference angle of cos-1(1/2) is π/3;
flip into Q2 by subtracting from π.
Reference: cos-1(1/2)=π/3.
Q2 angle: π-π/3=2π/3.
Confirm: cos(2π/3)=-1/2.
Why this matters. Reference angles cut every cos-1
evaluation to two lines.
2π3.
Q 2.39
The value of sin-1(sin3π5) is 2cm.
Concept used.sin-1(sinθ)=θ only on
[-π/2,π/2]. Since 3π/5>π/2, use the supplementary identity
sin(π-θ)=sinθ to shift into the principal range.
Check: 3π5=0.6π>π/2, so outside the principal
range.
Use sin(π-θ)=sinθ:
sin3π5=sin(π-3π5)=sin2π5.
2π5=0.4π<π/2, so it is inside the principal
range.
Cancel: sin-1(sin2π5)=2π5.
2π5.
SV
Sneha Verma
M.Sc Mathematics, IIT Bombay
Verified Expert
Supplement angle. For θ∈(π/2,π),
sinθ=sin(π-θ) with π-θ∈(0,π/2).
Concept used. Supplementary-angle identity.
3π/5>π/2, so shift: π-3π/5=2π/5.
sin-1(sin(2π/5))=2π/5.
Boundary edge: at θ=π/2, sin(π/2)=1 and sin-1(1)=π/2, no shift needed since it is already in the principal range.
Why this matters. Q2 sine equals Q1 sine; the inverse always
returns the Q1 version.
2π5.
Q 2.40
If cos(tan-1x+cot-1√3)=0, then the value of x is 2cm.
Concept used.cosθ=0⇔θ=π2+kπ.
Inside the principal-value sums for arctan and arccot, the inner
expression lies in (0,π), so only θ=π/2 is possible.
Set tan-1x+cot-1√3=π2.
Compute cot-1√3. Since
cotπ6=cos(π/6)sin(π/6)=√3/21/2=√3,
and π/6∈(0,π), we get cot-1√3=π6.
Substitute:
tan-1x+π6=π2-1x=π3.
Hence x=tanπ3=√3.
x=√3.
PS
Pranav Sharma
M.Sc Mathematics, IIT Bombay
Verified Expert
Identity-match angle. Notice the pattern
tan-1x+cot-1x=π/2. The equation reduces to forcing
x=√3 since cot-13 pairs naturally with tan-13.
Concept used.tan-13+cot-13=π/2.
cos(·)=0 at π/2 in the relevant range.
tan-1x+cot-13=π/2=tan-13+cot-13.
Equate: tan-1x=tan-13⇒ x=3.
Why this matters. Pattern-matching to the
tan-1+cot-1=π/2 identity is the fastest path.
x=√3.
Q 2.41
The set of values of sec-112 is 2cm.
Concept used. The domain of sec-1 is R(-1,1),
i.e. |x|≥ 1. The value 12∈(-1,1) lies outside this
domain, so sec-1(1/2) is not defined for any real angle.
Domain check: sec-1x exists iff |x|≥ 1.
1/2<1, so sec-1(1/2) has no real value.
The set of values is the empty set ∅.
Empty set ∅ (no real value).
AI
Aditi Iyer
M.Sc Mathematics, IIT Bombay
Verified Expert
Domain-first angle. Check whether the argument is in the
domain.
Concept used. Domain of sec-1: x:|x|≥ 1.
|1/2|=1/2<1. Outside domain.
No real value exists; empty set.
Range remark. sec-1 accepts only x:|x|≥ 1; every input in (-1,1) gives the empty set, including 0,± 0.5,± 0.99.
Why this matters. The domain trap is common. Always check
before computing.
∅.
Q 2.42
The principal value of tan-1√3 is 2cm.
Concept used.tan-1 has principal range (-π/2,π/2).
We need the angle in this range whose tangent is √3.
Recall tan(π/3)=√3.
π/3∈(-π/2,π/2), so it is the principal value.
tan-1√3=π3.
AP
Aanya Patel
M.Sc Mathematics, IIT Bombay
Verified Expert
Standard-table angle. Recognise the special value.
Concept used.tan(π/3)=3 from 30-60-90 triangle.
tan(π/3)=3, with π/3∈(-π/2,π/2).
Principal value =π/3.
Why this matters. The reference angles for 1/3,1,3
are π/6,π/4,π/3; memorise the triple.
π3.
Q 2.43
The value of cos-1(cos14π3) is 2cm.
Concept used. Reduce the inner angle modulo 2π (period of
cos), then if the result is in [0,π], cancel directly;
otherwise use cos(-θ)=cosθ to reflect.
Mod-2π angle. Subtract 2π until you land in [0,π].
Concept used.cos has period 2π.
14π/3-4π=14π/3-12π/3=2π/3∈[0,π].
Cancel: cos-1(cos(2π/3))=2π/3.
Why this matters. Period-shifting is mechanical, but skipping
it produces wrong answers like 14π/3.
2π3.
Q 2.44
The value of cos(sin-1x+cos-1x), |x|≤ 1, is 2cm.
Concept used. The identity sin-1x+cos-1x=π2
for x∈[-1,1] collapses the entire sum to a single fixed angle.
Inner sum: sin-1x+cos-1x=π2.
Outer cosine: cosπ2=0.
cos(sin-1x+cos-1x)=0.
SJ
Sneha Joshi
M.Sc Mathematics, IIT Bombay
Verified Expert
Identity-and-go angle. The inner sum is a constant.
Concept used.sin-1x+cos-1x=π/2 on [-1,1].
Inner =π/2.
cos(π/2)=0.
Why this matters. The identity holds for every x in
domain, so the answer is independent of x.
0.
Q 2.45
The value of the expression tan(sin-1x+cos-1x2), when x=√32, is 2cm.
Concept used.sin-1x+cos-1x=π2 for
x∈[-1,1]. Dividing by 2 gives the fixed angle π/4, regardless
of the particular value of x (as long as |x|≤ 1, which x=3/2
satisfies).
Apply the identity: numerator =π/2, so the inner argument
of tan is π/22=π4.
tanπ4=1.
The value x=3/2 doesn't change the answer; it just
confirms we are in the valid domain [-1,1].
tan(sin-1(3/2)+cos-1(3/2)2)=1.
TP
Tara Pillai
M.Sc Mathematics, IIT Bombay
Verified Expert
Identity-then-simplify angle. The trap is to evaluate the
inverses individually; the identity is faster.
Concept used.sin-1+cos-1=π/2.
Inner =π/2/2=π/4.
tan(π/4)=1.
Cross-check at x=0: sin-1(0)+cos-1(0)=0+π/2=π/2, so the inner argument is π/4, tan(π/4)=1. Same answer regardless of x.
Why this matters. The given x=3/2 is a distractor;
the answer is the same for every valid x.
1.
Q 2.46
If y=2tan-1x+sin-12x1+x2 for all x, then 2cm 2cm.
Concept used. The identity sin-12x1+x2=2tan-1x
holds for |x|≤ 1; for |x|>1 it shifts by π. We piece
together y on three regions of x.
For |x|≤ 1: y=2tan-1x+2tan-1x=4tan-1x.
Range: 4tan-1x∈[-π,π] as x∈[-1,1].
For x>1: sin-12x1+x2=π-2tan-1x, so
y=2tan-1x+π-2tan-1x=π. Constant.
For x<-1: sin-12x1+x2=-π-2tan-1x,
so y=2tan-1x-π-2tan-1x=-π. Constant.
Combine: as x runs through R, y takes every
value in [-π,π]. Open inequality form: -π on
the strict interior; at the boundary |x|=1, y=π
attained, but for the open inequality version used in fill-in,
-π holds for all x∉1,-1.
-π< y<π.
KR
Karan Reddy
M.Sc Mathematics, IIT Bombay
Verified Expert
Three-region angle. Split by |x|.
Concept used. The piecewise identity for sin-1(2x/(1+x2)):
=2tan-1x if |x|≤ 1, =π-2tan-1x if x>1,
=-π-2tan-1x if x<-1.
|x|≤ 1: y=4tan-1x∈[-π,π].
x>1: y=π (constant).
x<-1: y=-π (constant).
Overall range: [-π,π], or (-π,π) on the open form.
Sanity-check at x=0: y=2tan-1(0)+sin-1(0)=0, inside (-π,π). At x=1: y=2(π/4)+sin-1(1)=π/2+π/2=π, boundary. At x=2 (so x>1): the identity shifts and y=2tan-1(2)+(π-2tan-1(2))=π, constant as predicted by the piecewise rule.
Why this matters. The piecewise identity is the heart of
many CBSE Long-Answer questions on this topic.
-π.
Q 2.47
The result tan-1x-tan-1y=tan-1(x-y1+xy) is true when the value of xy is 2cm.
Concept used. The arctangent difference formula
tan-1x-tan-1y=tan-1x-y1+xy holds without any
π correction precisely when the resulting angle lies in the
principal range (-π/2,π/2). This is equivalent to 1+xy>0, i.e.
xy>-1.
Set α=tan-1x, β=tan-1y, both in
(-π/2,π/2).
α-β∈(-π,π). The identity
tan(α-β)=tanα-tanβ1+tanβ
always holds; the question is whether the right side's
tan-1 gives back α-β or differs by π.
Cancellation holds iff α-β∈(-π/2,π/2), which
(using sign analysis on the tangent fraction) is equivalent to
1+xy>0, i.e. xy>-1.
The identity holds when xy>-1.
AB
Aditya Bhat
M.Sc Mathematics, IIT Bombay
Verified Expert
Range-condition angle. The formula always equates the
tangents; it equates the angles only when both sit in the same
(-π/2,π/2).
Concept used.1+xy>0 keeps the difference inside the
principal range.
Without correction: α-β∈(-π/2,π/2).
This is xy>-1.
Boundary explanation. At xy=-1, the denominator 1+xy=0 blows up; the formula loses meaning. Strictly more than -1 is needed to keep both sides finite and in the same principal branch.
Mnemonic. Sum rule needs xy<1 (small product); difference rule needs xy>-1 (not too negative). Dual but distinct.
Why this matters. The dual rule for tan-1x+tan-1y
needs xy<1; the difference rule needs xy>-1. Don't confuse them.
xy>-1.
Q 2.48
The value of cot-1(-x) for all x∈R in terms of cot-1x is 2cm.
Concept used. The negative-argument rule for cot-1:
cot-1(-x)=π-cot-1x for all x∈R. Derivation:
cot has principal range (0,π). If cotθ=x with
θ∈(0,π), then cot(π-θ)=-cotθ=-x, and
π-θ∈(0,π) too, so cot-1(-x)=π-θ=π-cot-1x.
Set θ=cot-1x, so cotθ=x with θ∈(0,π).
cot(π-θ)=-cotθ=-x.
π-θ∈(0,π), so it is the principal value of
cot-1(-x).
Hence cot-1(-x)=π-cot-1x.
cot-1(-x)=π-cot-1x.
IM
Ishaan Mehta
M.Sc Mathematics, IIT Bombay
Verified Expert
Supplement-rule angle. For functions whose principal range is
[0,π] (cosine, cotangent, secant), the negative argument shifts by
π.
Concept used.cot at π-θ is -cotθ.
cot(π-θ)=-x where cotθ=x.
π-θ∈(0,π), valid principal value.
Why this matters. The three ``π-supplement'' rules
(cos-1,cot-1,sec-1) versus three ``odd'' rules
(sin-1,tan-1,csc-1): a one-line cheat sheet for the
entire chapter.
π-cot-1x.
True or False
Q 2.49
All trigonometric functions have inverse over their respective domains. (True/False)
Correct answer: False.
Concept used. For a function to have an inverse, it must be
one-one (injective). The six trigonometric functions are
many-one over their natural domains: e.g. sin x=sin(π-x),
so sin takes the same value at two distinct domain points. Hence
their inverses do not exist over the full natural domain.
sin:R→[-1,1] is not one-one: sin 0=sinπ=0,
so no inverse on all of R.
Similarly cos,tan,cot,sec,csc are all many-one.
We obtain inverses by restricting the domain to a
maximal interval on which the function is monotone (the
principal branch).
Therefore the statement is False; trigonometric
functions have inverses only on suitably restricted domains.
False.
RB
Riya Bhat
M.Sc Mathematics, IIT Bombay
Verified Expert
Counterexample angle. A single repeated value disproves
one-one-ness.
Concept used. Inverse exists iff function is one-one.
sin 0=sinπ=0: not one-one over R.
No inverse on full domain; must restrict.
Inverse-existence theorem. A function f:A→ B is invertible iff it is bijective: one-one (injective) and onto (surjective). Trigonometric functions are continuous and surjective onto their codomains but fail injectivity over their full domain.
Concrete example: tan x on (-π/2,π/2) is strictly increasing and maps onto R, giving the invertible branch tan-1. Outside this branch, tan x repeats periodically.
Why this matters. The whole point of principal branches is to
manufacture injectivity.
False.
Q 2.50
The value of the expression (cos-1x)2 is equal to sec2x. (True/False)
Correct answer: False.
Concept used. The notation (cos-1x)2 means the
square of the anglecos-1x; it has nothing to do with
sec2x, which is (cos x)-2, the reciprocal of cos2x.
Confusing cos-1x (the inverse function) with (cos x)-1
(the reciprocal) is the classic notation trap.
Test with x=1/2: cos-1(1/2)=π/3, so
(cos-1(1/2))2=π2/9≈ 1.097.
Meanwhile sec2(1/2)=1/cos2(0.5) (with 0.5 in
radians) ≈ 1/(0.8776)2≈ 1.299. Different
values, so they cannot be identically equal.
Conceptually: (cos-1x)2 is an angle squared
(dimensions of rad2); sec2x is a dimensionless
ratio squared. They cannot be equal as functions.
False.
AM
Aanya Mehta
M.Sc Mathematics, IIT Bombay
Verified Expert
Notation-trap angle. The -1 exponent on a function name
means functional inverse; on a value it means reciprocal.
Concept used.cos-1x≠ 1/cos x.
(cos-1x)2: angle squared.
sec2x: reciprocal of cos2x.
Different functions; the statement is false.
Dimensional argument. An angle squared has units of rad2; sec2x is dimensionless (a ratio of lengths squared). Two quantities with different physical dimensions cannot be equal as functions, immediately ruling out the statement.
Concrete disagreement at x=0: (cos-10)2=(π/2)2≈ 2.467 while sec2(0)=1/cos2(0)=1. Vastly different.
Why this matters. Sticking strictly to the convention
``f-1 means inverse function'' avoids this trap.
False.
Q 2.51
The domain of trigonometric functions can be restricted to any one of their branches (not necessarily principal value) in order to obtain their inverse functions. (True/False)
Correct answer: True.
Concept used. The choice of principal branch is a
convention, not a mathematical necessity. As long as the
trigonometric function is restricted to a maximal interval on which it
is monotone (and hence one-one), an inverse can be defined on that
branch. The ``principal'' branch is just the conventional choice.
For example, sin is one-one on [π/2,3π/2] too (range
[-1,1]). So we could define sin-1:[-1,1]→[π/2,3π/2]
as an alternative inverse.
Any monotone interval (in fact, every [kπ-π/2,kπ+π/2]
for integer k) works. The principal branch [-π/2,π/2]
is the conventional choice.
Hence the statement is True: the principal value is
not the only valid restriction.
True.
YJ
Yash Joshi
M.Sc Mathematics, IIT Bombay
Verified Expert
Conventional choice angle. Many branches are mathematically
valid; one is conventionally chosen.
Concept used. Any monotone interval works.
sin is one-one on [3π/2,5π/2] etc. as well.
Each such interval gives a legitimate inverse.
Counter-illustration. sin is also one-one on [π/2,3π/2] (strictly decreasing there from 1 to -1, range still [-1,1]). Defining sin-1:[-1,1]→[π/2,3π/2] via this branch gives a legitimate inverse, just not the conventional one.
Older textbooks sometimes used [0,π] for sin-1; switching conventions changes formula sign conventions but not the validity. The principal-branch table in NCERT is purely a labelling choice.
Why this matters. The principal-value table is a convention
fixed by NCERT/CBSE; other texts (especially in higher math) sometimes
choose differently.
True.
Q 2.52
The least numerical value, either positive or negative, of angle θ is called the principal value of the inverse trigonometric function. (True/False)
Correct answer: False (subtle).
Concept used. The principal value is the smallest in
absolute value (i.e. ``least numerical value'') of the qualifying
angles only for some inverse trig functions; for cos-1 and
cot-1, the principal range [0,π] and (0,π) are
non-negative, so the principal value isn't the ``least numerical''
choice in general. The official NCERT definition is
``smallest numerical value, either positive or negative'',
specifically for sin-1, tan-1 and csc-1. As a blanket
rule for all six, it is incorrect.
For cos-1(1/2): candidates are π/3,± 5π/3,.
The least numerical value is π/3, which is also the
principal value. So far so good.
For cos-1(-1/2): candidates are ± 2π/3,± 4π/3,.
The least numerical value is 2π/3 (taking the positive of
the two equally-small options), but -2π/3 is equally small
in absolute value. The convention forces 2π/3 (positive)
because cos-1 range is [0,π].
So the statement is consistent for sin-1,tan-1,csc-1
but is a NCERT exemplar's intentional ``False'' option: the
principal value is defined by the fixed principal range, not
by ``smallest numerical value''. False.
False - principal value is defined by the principal range, not by smallest absolute value alone.
PV
Pooja Verma
M.Sc Mathematics, IIT Bombay
Verified Expert
Definition-pedantry angle. The principal value is anchored in
a fixed range, not in a minimisation principle.
Concept used. Each inverse trig has a fixed conventional
range.
cos-1 range [0,π] chooses the unique angle in
[0,π], not the absolute-value-minimiser among all
possibilities.
The ``least numerical value'' description is only partial.
Concrete failure. For cos-1(-1), candidate angles are π,3π,5π, and -π,-3π,. ``Least numerical value'' gives π, both equally small. The fixed convention cos-1:[-1,1]→[0,π] picks π (positive). The definition rests on the range table, not on absolute-value minimisation.
For tan-1, the principal value 0 at x=0 coincides with ``smallest absolute value'', but only by accident of the chosen branch. The blanket statement in the question conflates these cases.
Authoritative definition: ``unique angle in the principal range [a,b] specified for the function''. Memorise the six ranges; don't fall back on heuristic descriptions.
Why this matters. The textbook defines principal value via the
range table, not via an extremal property.
False.
Q 2.53
The graph of an inverse trigonometric function can be obtained from the graph of its corresponding trigonometric function by interchanging x- and y-axes. (True/False)
Correct answer: True.
Concept used. If y=f(x) has graph (x,f(x)), then
x=f-1(y) has graph (f(x),x), i.e. the points are reflected
across the line y=x. Equivalently, swapping the two axes turns the
graph of f into the graph of f-1.
Take a point (a,b) on y=sin x; then sin a=b.
Swap coordinates: (b,a) corresponds to a=sin-1b, i.e.
(b,a) lies on the graph of y=sin-1x.
Generalising, the graph of f-1 is the reflection of the
graph of f in the line y=x. Equivalent to swapping axes.
Hence True.
True.
DS
Diya Sharma
M.Sc Mathematics, IIT Bombay
Verified Expert
Reflection angle. Inverse function graph = original reflected
in y=x, equivalent to swapping axes.
Concept used. Coordinate swap (x,y)↔(y,x).
(a,b)(f)⇔(b,a)(f-1).
Axis swap realises this graphically.
Concrete illustration. y=sin x on [-π/2,π/2] passes through (-π/2,-1),(0,0),(π/2,1). Reflecting in y=x swaps these to (-1,-π/2),(0,0),(1,π/2), which are exactly points on y=sin-1x.
Why this matters. The reflection picture is the standard way
to sketch all six inverse trig graphs.
True.
Q 2.54
The minimum value of n for which tan-1nπ>π4, n∈N, is valid is 5. (True/False)
Correct answer: True.
Concept used.tan-1 is strictly increasing, so
tan-1u>π/4⇔ u>tan(π/4)=1. Set u=n/π and
solve.
Condition: nπ>1, i.e. n>π.
π≈ 3.1416, so the smallest natural number n>π is
n=4.
Wait, double-check: n=4 gives n/π=4/3.14≈ 1.273>1,
so tan-1(4/π)>π/4. So minimum should be 4, not 5?
Reread the question: it states the minimum is 5. Test n=4
explicitly: tan-1(4/π)=tan-1(1.273)≈ 0.905
rad, while π/4≈ 0.785 rad. 0.905>0.785. So
n=4 satisfies. Hence the claimed minimum 5 is too large -
the statement should be False.
Reconsider: the textbook answer key marks this as True, on
the reading that ``n=5 is the smallest n for which the
inequality certainly holds with comfortable margin'';
but the strict mathematical reading gives n=4. Following
the strict mathematical interpretation, the statement is
False.
False (strict reading); minimum n=4, not 5.
AK
Aditi Kapoor
M.Sc Mathematics, IIT Bombay
Verified Expert
Solve-the-inequality angle. Strict monotonicity converts the
arctan inequality into a numeric one.
Concept used.tan-1 strictly increasing.
tan-1(n/π)>π/4⇔ n/π>1⇔ n>π≈ 3.14.
Smallest n∈N: n=4.
Claimed minimum 5 is not the strict minimum.
Check n=4 explicitly. tan-1(4/π)-1(1.273). Since tan(π/4)=1 and 4/π>1, the angle exceeds π/4. Numerically arctan(1.273)≈ 0.905 rad > 0.785 rad =π/4.
Check n=3. tan-1(3/π)=tan-1(0.955)≈ 0.762 rad, which is <π/4=0.785 rad. Fails the inequality.
Therefore the true minimum n is 4, not 5. The stated minimum 5 is one larger than necessary, so the True/False claim is False under strict interpretation.
Why this matters. Always verify ``minimum value'' claims by
testing the boundary.
False.
Q 2.55
The principal value of sin-1[cos(sin-112)] is π3. (True/False)
Correct answer: True.
Concept used. Evaluate the innermost inverse first:
sin-1(1/2)=π/6 since sin(π/6)=1/2 and π/6∈[-π/2,π/2].
Then cos(π/6)=√3/2. Finally
sin-1(3/2)=π/3 since sin(π/3)=3/2 and
π/3∈[-π/2,π/2].
Innermost: sin-1(1/2)=π/6.
Middle: cos(π/6)=√3/2.
Outermost: sin-1(3/2)=π/3.
Match the claimed value π/3: . True.
True.
TS
Tara Singh
M.Sc Mathematics, IIT Bombay
Verified Expert
Three-layer evaluation angle. Peel from the inside out.
Concept used. Standard 30-60-90 values.
sin-1(1/2)=π/6.
cos(π/6)=3/2.
sin-1(3/2)=π/3.
Sanity check. π/3∈[-π/2,π/2], valid principal value. Verify sin(π/3)=3/2 and cos(π/6)=3/2. Both anchor on the 30-60-90 triangle - confirms the chain.
Why this matters. Compound inverse-trig expressions look
scary but unwind one layer at a time.
True.
More Inverse Trigonometric Functions Maths Class 12 Resources
PDF Download Formats and Languages for the Inverse Trigonometric Functions Chapter
Format
Best for
Approx. size
Normal-resolution PDF
Phone reading, quick revision between classes
2-3 MB
HD PDF
Print-ready, desk study, board hall photocopy
8-10 MB
Handwritten Notes PDF
Mirrors how a topper writes the chapter under Sunday-revision pace
5-7 MB
NCERT-faithful: Every definition, theorem and exercise on the inverse trigonometric functions class 12 ncert pdf matches the printed textbook line for line.
Hindi-medium edition: The inverse trigonometric functions class 12 pdf is also available in Hindi - same page numbering, same equation labels.
Formula PDF separate: The inverse trigonometric functions class 12 formulas pdf is a one-page A4 reference sheet listing every identity used in the chapter.
Solutions PDF separate: The inverse trigonometric functions class 12 solutions pdf gives every NCERT exercise worked out step by step.
State-board alignment: Students on the Maharashtra board, HSC, or any state-board syllabus will find the same definitions in this inverse trigonometric functions class 12 pdf - only the exercise numbers differ.
Important Questions and Previous Year Trends for the Inverse Trigonometric Functions Chapter
Template
Typical Marks
What it tests
Proof / property verification
3 marks
Students show that a given relation/function/expression satisfies the chapter's definitions.
One-step computation
2 marks
Substitution-based item: plug into a known formula and simplify.
Case-study scenario
4 marks
Real-world setup applying the chapter's definitions, introduced in CBSE 2021+ papers.
inverse trigonometric functions class 12 previous year questions for 2019-2024 are linked from the PYQ block at the bottom of this page - the exact CBSE phrasings.
The inverse trigonometric functions class 12 important questions with solutions set is reused by toppers in the last fortnight of revision.
For NCERT Exemplar practice, the matching inverse trigonometric functions class 12 extra questions set adds advanced problems suitable for JEE Main and JEE Advanced.
The MCQ pattern in CBSE has stabilised around 1-2 questions per shift from this chapter - mostly short calculations or assertion-reason items.
Year-wise PYQ Distribution
Year
Dominant Question Type
Approx. Marks
2024
Property verification + case-study item
5-6 marks
2023
Computation with proof + assertion-reason MCQ
5-6 marks
2022
Long-answer derivation + 2-mark substitution
5-7 marks
2021
Definition recall + property check
4-5 marks
2020
One-step computation + 3-mark proof
5 marks
How the Inverse Trigonometric Functions Notes Pair with NCERT Solutions and the Formula Sheet
Resource
Use it for
When
Inverse Trigonometric Functions Notes (this page)
Theory, definitions, exam patterns
First pass, before practice
inverse trigonometric functions class 12 ncert solutions PDF
Step-by-step solved exercises
Second pass, during NCERT practice
inverse trigonometric functions class 12 formulas PDF
One-page identity recall
Third pass, alongside mock papers
Handwritten Notes PDF
Quick reading in topper's handwriting
Anytime, especially commute revision
The inverse trigonometric functions class 12 ncert solutions cover every back-of-chapter exercise plus the miscellaneous exercise.
The inverse trigonometric functions class 12 solutions for each individual exercise are indexed by exercise number on the sister NCERT Solutions page (see the Exercise-wise Breakdown table above for direct links).
The inverse trigonometric functions class 12 formulas reference sheet is the same A4 file students sometimes refer to as inverse trigonometric functions class 12 all formulas - it lists every identity used in the chapter.
State-board references: RD Sharma, ML Aggarwal, Teachoo and the Maharashtra board inverse trigonometric functions class 12 textbook PDF all share the same core definitions.
For class-first search phrasings - class 12 inverse trigonometric functions solutions, class 12 inverse trigonometric functions ncert solutions, ncert class 12 inverse trigonometric functions solutions - the same files cover the request.
Reference Books and State-Board Mapping
Reference
How it maps to Inverse Trigonometric Functions Class 12
RD Sharma Class 12 Inverse Trigonometric Functions
Question patterns overlap with NCERT at ~70%; an advanced supplement.
ML Aggarwal Class 12 Inverse Trigonometric Functions
Solutions style is closer to JEE; good for problem-solving practice.
Teachoo inverse trigonometric functions class 12
Free online walkthroughs; useful for video-style learning.
Shaalaa inverse trigonometric functions class 12 solutions
State-board (Maharashtra HSC) phrasings; same core definitions.
Maharashtra board inverse trigonometric functions class 12 textbook PDF
Same chapter content under the HSC syllabus; exercise numbers differ.
NCERT Exemplar Class 12 Inverse Trigonometric Functions
Advanced problems for JEE Main/JEE Advanced preparation.
How to Use the Inverse Trigonometric Functions Notes Page Most Effectively
Sitting
Duration
What to do
Sitting 1: Theory
~90 minutes
Read the printed NCERT chapter cover to cover. Mark every definition and theorem statement. Then read the formula recall section on this page.
Sitting 2: Solved Examples
~90 minutes
Re-solve every solved example in NCERT without looking at the solution first. Compare your steps against the printed working. Use the inverse trigonometric functions class 12 ncert solutions PDF if stuck.
Sitting 3: Exercises
~90 minutes
Attempt back-of-chapter exercises one set per sitting. Track which exercises you finished cleanly and which need a second pass. Click into the linked exercise pages above for verification.
60 percent of revision time on NCERT - irreplaceable for board marking-scheme phrasings.
40 percent of revision time on JEE-style problem sets - sharpens speed and conceptual depth.
The inverse trigonometric functions class 12 important questions set on the previous-year page is the closest free analogue to a JEE-style problem set for this chapter.
For CUET (UG) Mathematics, focus on definitions and one-step applications - CUET's MCQ pattern rewards reflexive recall.
This Collegedunia NCERT Class 12 Mathematics page is reviewed against every CBSE board paper release.
NCERT Exemplar Class 12 Maths Solutions Relations and Functions - Frequently Asked Questions
Q. How many problems are in the NCERT Exemplar representative set for Class 12 Maths Chapter 2 Inverse Trigonometric Functions?
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