Mathematics Content Strategist | Olympiad Coach, 10 Years | Updated on - May 24, 2026
The Inverse Trigonometric Functions Class 12 NCERT Solutions page compiles NCERT Class 12 Mathematics Chapter 2 into a download-ready resource, aligned to the 2026-27 NCERT syllabus. The page covers definitions, solved examples, exam-weightage data and common mistakes, every formula matched to the CBSE marking scheme.
The solutions PDF covers all parts of every question and writes each step in full. The file is part of the NCERT Solutions Class 12 Maths library and follow the notation of the official textbook.
CBSE Weightage: 5-7 marks (Relations and Functions + Inverse Trig chapter cluster)
JEE Main Weightage: 3-5% (1-2 questions on principal values / identity simplification)
Why this exercise carries disproportionate weight: Inverse trig appears in three downstream chapters (Continuity and Differentiability, Integrals, Differential Equations). Every identity you internalise here pays compound interest across the rest of the syllabus. CBSE knows this, which is why one full short-answer slot in every board paper since 2018 has come from these 17 problems.
These solutions were prepared by Collegedunia's senior Mathematics faculty against the 2026-27 NCERT print, with every principal-value justification spelled out, every identity step named, and every conversion between sin-1, cos-1, tan-1 verified. The working follows CBSE marking-scheme conventions, so the format mirrors what examiners reward.
Concepts Tested in the Class 12 Maths Chapter 2 Miscellaneous Exercise
The Inverse Trigonometric Functions Class 12 NCERT Solutions address this in the same order as the NCERT textbook.
The 17 problems cluster into four concept families. The table below maps each question to its underlying technique so you can revise the right identity before attempting the problem.
Question Cluster
Underlying Concept
NCERT Section
Marks Range (CBSE)
Q1, Q2
Principal value evaluation of sin-1, cos-1, tan-1
2.2 Basic Concepts
1-2
Q3 to Q5
Prove identities using 2tan-1x and double-angle forms
2.3 Properties
3-4
Q6 to Q10
Simplify nested inverse trig expressions, substitute x = tanθ etc.
2.3 Properties
4-5
Q11 to Q14
Solve equations involving tan-1 sums and differences
2.3 Properties
4-6
Q15 to Q17
MCQs on principal value and identity-application
All sections
1-2 (MCQ)
Notice the heavy concentration on Section 2.3 properties. That is the revision priority: the tan-1x + tan-1y formula, the double-angle conversions, and the substitution tricks (x = tanθ , x = cos 2θ ) account for roughly 11 of the 17 problems.
Inverse Trigonometric Functions Miscellaneous Video Walkthrough
Previous Year Question Trends - Inverse Trigonometric Functions Chapter
The table below summarises the recent CBSE Class 12 pattern for this chapter and is a quick pre-exam reference.
Year
CBSE Marks
Question Type
Source Pattern
2025
4
Prove identity using tan-1 addition
Modelled on Misc Q8
2024
5
Solve tan-1 equation
Modelled on Misc Q13, Q14
2023
4
Principal value + simplification
Modelled on Misc Q1, Q2
2022
5
Prove sin-1 + cos-1 identity
Modelled on Misc Q5, Q6
2021
4
Substitution trick x = tanθ
Modelled on Misc Q9, Q11
Common Mistakes Students Make in the Miscellaneous Exercise
Mistake 1 - Ignoring the xy < 1 restriction on tan-1 addition. The identity tan-1x + tan-1y = tan-1(x+y1-xy) holds only when xy < 1 . When xy > 1 , an additive π (or -π ) is required. Q8 catches students who forget this; the π/4 result depends on pairing fractions carefully.
Mistake 2 - Returning an angle outside the principal-value range. In Q1 and Q2, cos-1(cosθ) is not always θ ; it equals θ only if θ ∈ [0, π] . Otherwise reduce first. Marking schemes deduct 1 full mark for an answer outside the principal range.
Mistake 3 - Forgetting the substitution justification. Q9 and Q11 require x = tan2θ or x = cos 2θ . Students who jump straight to the substituted form without stating "let x = tan2θ " lose a full method mark. Always declare the substitution before using it.
Quick Revision Tips Before You Attempt the Miscellaneous Exercise
Tip 1: Memorise the six principal-value ranges ( sin-1 ∈ [-π/2, π/2] , cos-1 ∈ [0, π] , tan-1 ∈ (-π/2, π/2) , etc.). Half of all errors in Q1, Q2, Q15, Q16 trace back to a wrong range.
Tip 2: Convert every sin-1 and cos-1 in the question to a tan-1 form before applying the addition rule. The tan-1 addition formula is the most useful identity in this exercise.
Tip 3: For substitution problems (Q9, Q10, Q11), the trigger words are "√1+x" or "√1-x2" or nested radicals. Reach immediately for x = tanθ , x = cos 2θ , or x = tan2θ . The substitution collapses the expression in two lines.
PDF Download Formats and Languages for the Inverse Trigonometric Functions Chapter
The table below summarises the recent CBSE Class 12 pattern for this chapter and is a quick pre-exam reference.
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
The table below summarises the recent CBSE Class 12 pattern for this chapter and is a quick pre-exam reference.
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
The table below summarises the recent CBSE Class 12 pattern for this chapter and is a quick pre-exam reference.
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.
All NCERT Solutions for Inverse Trigonometric Functions Misc with Step-by-Step Working
Every NCERT textbook question for Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions Misc 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.
Questions
Q 2.1
Find the value of cos-1(cos13π6).
Concept used. The identity cos-1(cos u) = u holds only
when u ∈ [0,π]. The cosine function has period 2π and is
even, so cos u = cos(u - 2π k) = cos(2π - u). We use these
to fold the inner angle into the principal range.
Check.13π6 = 2π + π6, so
13π6 > 2π > π, well outside the principal range [0,π].
Quick reading.13π6 is one full revolution past
π6. So the cosine is the same as cos(π/6) and the
principal answer is π/6.
13π6 - 2π = π6.
cos(π/6) ∈ [0,π], so cos-1(cos(π/6)) = π/6.
Why this matters. The shortcut "subtract 2π k first, then
adjust within [0,2π)" handles every cos-1(cos u) regardless
of how big u is.
π6.
Q 2.2
Find the value of tan-1(tan7π6).
Concept used. The identity tan-1(tan u) = u holds only
when u ∈ (-π/2, π/2). Tangent has period π, so
tan u = tan(u - π k) for any integer k. Subtract multiples of
π to shift u into the principal open interval.
Check.7π6 > π2 = 3π6, so
outside (-π/2,π/2).
Subtract π.tan(u - π) = tan u gives
tan(7π6) = tan(7π6 - π)
= tan(7π - 6π6) = tan(π6).
Check.π6 ∈ (-π/2, π/2) .
Apply tan-1.tan-1(tan7π6) = tan-1(tanπ6) = π6.
tan-1(tan7π6) = π6.
VR
Vivaan Reddy
M.Tech CS, IIT Madras
Verified Expert
Strategic angle. Tangent has period π, so subtracting one
full π from 7π6 drops us inside the open principal
interval. Confirm by computing the tangent value.
7π6 - π = π6, which lies in (-π/2,π/2).
tan(π/6) = 1/√3, and so does tan(7π/6) .
tan-1(tan(7π/6)) = tan-1(tan(π/6)) = π/6.
Why this matters. Tangent's π-period (not 2π) is the
single fact that makes tan-1(tan u) different from
sin-1(sin u). Use π-shifts, not 2π-shifts.
π6.
Q 2.3
Prove that 2sin-135 = tan-1247.
Concept used. The doubling identity
2sin-1x = sin-1(2x√1-x2) holds for
|x| ≤ 1√2, but a cleaner approach here uses
2sin-1x = tan-1(2x√1-x21 - 2x2)
when the right-hand denominator is positive. Equivalently, set
θ = sin-1(3/5) and compute tan 2θ via the
double-angle formula
tan 2θ = 2tanθ1 - tan2θ.
Set θ. Let θ = sin-1(3/5), so
sinθ = 3/5 and θ ∈ (0,π/2) since 3/5 > 0.
Build the right triangle. With sinθ = 3/5
in (0,π/2): opposite =3, hypotenuse =5, so adjacent
=√52 - 32 = √16 = 4. Hence cosθ = 4/5
and tanθ = 3/4.
Compute tan 2θ. Apply the double-angle
formula:
tan 2θ = 2tanθ1 - tan2θ
= 2 · 341 - 916
= 32716.
Simplify.3/27/16 = 32 · 167
= 4814 = 247.
Confirm 2θ is in the principal range of tan-1.
From sinθ = 3/5 we have θ < π/4 (since
sin(π/4) = 1/√2 ≈ 0.707 > 0.6 = 3/5 is false;
actually 1/√2 > 3/5, so θ < π/4 holds when
sinθ < sin(π/4)). Hence
2θ < π/2, so 2θ ∈ (0, π/2) ⊂ (-π/2, π/2).
Therefore tan-1(tan 2θ) = 2θ.
Conclude.2θ = tan-1(24/7), i.e.
2sin-1(3/5) = tan-1(24/7).
2sin-135 = tan-1247.
PN
Pranav Nair
Ph.D Mathematics, IIT Delhi
Verified Expert
Picture-first. Sketch the right triangle that encodes
sin-1(3/5), read off tanθ = 3/4, then apply the
double-angle formula for tangent.
Since θ < π/4 (because 3/5 < 1/√2),
2θ < π/2 lies in the principal range, so
tan-1(tan 2θ) = 2θ.
Why this matters. Identities of the form
"2sin-1(a/c) = tan-1(?)" all yield to the same recipe:
build the right triangle, read tanθ, apply the double-angle
formula.
2sin-1(3/5) = tan-1(24/7).
Q 2.4
Prove that sin-1817 + sin-135 = tan-17736.
Concept used. Let A = sin-1(8/17) and B = sin-1(3/5),
both in (0,π/2). Read tan A and tan B off right triangles
(8,15,17) and (3,4,5), then apply the tangent sum formula and
finally take tan-1.
Build triangle for A.sin A = 8/17 with hypotenuse 17
and opposite 8 gives adjacent √172 - 82
= √289 - 64 = √225 = 15. So cos A = 15/17 and
tan A = 8/15.
Build triangle for B.sin B = 3/5 gives
cos B = 4/5 and tan B = 3/4 (the standard 3,4,5 triangle).
Apply tangent sum.tan(A + B) = tan A + tan B1 - tan A tan B
= 815 + 341 - 815 · 34.
Apply tan-1. Both A and B are in (0,π/2)
with tan A < 1 and tan B < 1, so tan A tan B < 1;
hence A + B < π/2 and the tangent of the sum is positive.
A + B ∈ (0,π/2) lies in the principal range, so
tan-1(tan(A+B)) = A + B, giving
A + B = tan-1(77/36).
sin-1817 + sin-135 = tan-17736.
AB
Aanya Banerjee
M.Sc Applied Mathematics, IIT Kanpur
Verified Expert
Picture-first. Two Pythagorean triples (8,15,17) and (3,4,5)
encode the two inverse sines. The tangent of the sum follows directly
from their tangent ratios.
Why this matters. Memorising the first few Pythagorean triples
(3,4,5), (5,12,13), (8,15,17), (7,24,25) makes the right
triangles immediate and turns these proofs into one-line tangent-sum
computations.
tan-1(77/36).
Q 2.5
Prove that cos-145 + cos-11213 = cos-13365.
Concept used. Let A = cos-1(4/5) and B = cos-1(12/13),
both in [0,π/2]. Then cos(A+B) follows from the cosine sum
formula cos(A+B) = cos A cos B - sin A sin B. Read sin A and
sin B off Pythagorean triples.
For A:cos A = 4/5, hence by
sin2A + cos2A = 1,
sin A = √1 - (4/5)2 = √1 - 16/25 = √9/25 = 3/5
(positive since A ∈ [0,π/2]).
For B:cos B = 12/13, hence
sin B = √1 - 144/169 = √25/169 = 5/13.
Cosine sum formula.cos(A+B) = cos A cos B - sin A sin B
= 45 · 1213 - 35 · 513.
Compute the two products.45 · 1213 = 4865;
35 · 513 = 1565.
Subtract.cos(A+B) = 4865 - 1565 = 3365.
Apply cos-1. Both A and B are in [0,π/2], so
A + B ∈ [0,π], exactly the principal range of cos-1.
Hence cos-1(cos(A+B)) = A + B, giving
A + B = cos-1(33/65).
cos-145 + cos-11213 = cos-13365.
RP
Riya Patel
B.Tech CSE, IIT Roorkee
Verified Expert
Structural observation. The two given cosines come from the
triples (3,4,5) and (5,12,13). Read off the sines and plug into the
cosine-sum formula. Three multiplications and a subtraction give the
answer.
sin A = 3/5, sin B = 5/13 (from Pythagoras).
cos A cos B = (4/5)(12/13) = 48/65.
sin A sin B = (3/5)(5/13) = 15/65.
cos(A+B) = 48/65 - 15/65 = 33/65.
A + B ∈ [0,π], so A+B = cos-1(33/65).
Why this matters. Pairs of cosines built from neighbouring
Pythagorean triples almost always combine cleanly via the cosine
addition formula. Expect a tidy fraction (here 33/65).
cos-1(33/65).
Q 2.6
Prove that cos-11213 + sin-135 = sin-15665.
Concept used. Let A = cos-1(12/13) and
B = sin-1(3/5), both in [0,π/2]. We compute sin(A+B) via
the sine sum formula
sin(A+B) = sin A cos B + cos A sin B.
For A:cos A = 12/13, so
sin A = √1 - 144/169 = 5/13 (positive in [0,π/2]).
For B:sin B = 3/5, so
cos B = √1 - 9/25 = 4/5 (positive in [0,π/2]).
Sine sum formula.sin(A+B) = sin A cos B + cos A sin B
= 513 · 45 + 1213 · 35.
Compute the two products.513 · 45 = 2065;
1213 · 35 = 3665.
Add.sin(A+B) = 2065 + 3665 = 5665.
Apply sin-1. We need A + B ∈ [-π/2, π/2].
Both A, B ∈ [0,π/2], so A + B ∈ [0,π]. We must
check A + B ≤ π/2. Equivalently, sin(A+B) ≤ 1,
which is automatic, but we need a stronger check:
A + B should be in [0,π/2]. Since sin A = 5/13 < 4/5 = cos B,
we have sin A < sin(π/2 - B) = cos B, so
A < π/2 - B, i.e. A + B < π/2. Hence A + B ∈ (0,π/2),
which lies inside the principal range of sin-1, so
sin-1(sin(A+B)) = A + B and the proof closes:
A + B = sin-1(56/65).
cos-11213 + sin-135 = sin-15665.
KV
Karan Verma
M.Sc Mathematics, ISI Kolkata
Verified Expert
Strategic angle. Convert one inverse to match the other
function. Here both belong to first-quadrant triangles, so reading
all four ratios takes one minute.
Triple (5,12,13): gives both cos A = 12/13 and sin A = 5/13.
Triple (3,4,5): gives both sin B = 3/5 and cos B = 4/5.
Check A + B < π/2 so the result is in the sin-1 principal
range, then A + B = sin-1(56/65).
Why this matters. Mixing sin-1 and cos-1 in the
same sum identity is the most common NCERT trick. Always pick the
target trig (sin vs cos) so that the cross terms come for free.
sin-1(56/65).
Q 2.7
Prove that tan-16316 = sin-1513 + cos-135.
Concept used. Let A = sin-1(5/13) and B = cos-1(3/5),
both in [0,π/2]. Read tan A and tan B off the triples
(5,12,13) and (3,4,5), apply the tangent sum formula, and take
tan-1.
For A:sin A = 5/13, cos A = 12/13 (triple (5,12,13)),
so tan A = 5/12.
For B:cos B = 3/5, sin B = 4/5 (triple (3,4,5)),
so tan B = 4/3.
Apply tangent sum.tan(A+B) = tan A + tan B1 - tan A tan B
= 512 + 431 - 512 · 43.
Divide.tan(A+B) = 21/124/9 = 2112 · 94
= 18948 = 6316
(dividing numerator and denominator by 3).
Check range.A + B exceeds π/2? We have
sin A = 5/13 ≈ 0.385 and cos B = 3/5 = 0.6, so
sin A < cos B = sin(π/2 - B), hence A < π/2 - B,
so A + B < π/2. Therefore A + B ∈ (-π/2,π/2),
and tan-1(tan(A+B)) = A + B, giving
A + B = tan-1(63/16).
tan-16316 = sin-1513 + cos-135.
AS
Aditya Singh
M.Tech CS, IIT Madras
Verified Expert
Structural observation. Convert both inverses to tangents,
apply the sum formula, simplify the resulting fraction.
A + B ∈ (0,π/2) lies in the principal range of tan-1,
so A + B = tan-1(63/16).
Why this matters. Pythagorean triples appear so often in this
exercise because they make every ratio a nice rational number. Build a
mental table of the first half-dozen triples.
tan-1(63/16).
Q 2.8
Prove that tan-1x = 12cos-1(1-x1+x), x ∈ [0,1].
Concept used. The substitution x = tan2(θ/2) or,
better, x = tanθ with θ ∈ [0,π/4], turns the right
side into a clean expression via the double-angle formula
cos 2θ = 1 - tan2θ1 + tan2θ.
Substitute. Let x = tanθ with
θ ∈ [0,π/4], which is the range of tan-1 on
[0,1]. Then θ = tan-1x and 2θ ∈ [0,π/2].
Rewrite the inner fraction. The aim is to show that
1-x1+x equals cos 2θ:
cos 2θ = 1 - tan2θ1 + tan2θ.
But the problem gives 1-x1+x = 1 - tanθ1 + tanθ,
not 1 - tan2θ1 + tan2θ. So a fresh
substitution is needed.
Better substitution. Put x = cos 2φ with
2φ ∈ [0,π/2] (since x ∈ [0,1] corresponds to
2φ ∈ [0,π/2], i.e. φ ∈ [0,π/4]). Then
cos-1x = 2φ, so the right-hand side
12cos-1(1-x1+x) requires us
to first compute 1-x1+x in terms of φ:
1-x1+x = 1 - cos 2φ1 + cos 2φ
= 2sin2φ2cos2φ = tan2φ.
Apply cos-1 on the RHS. The expression becomes
cos-1(tan2φ), which is not obviously clean. So we
rethink: use instead x = tan2t. Better yet, follow the NCERT
hint and put x = cos 2θ (so 2θ = cos-1x).
Then we need to identify the angle whose cosine is
1-x1+x = tan2θ. That is not directly a cosine.
We therefore replace the hint with the simpler chain:
prove the equivalent identity
2tan-1x = cos-1(1-x21+x2)
for x ≥ 0, and then substitute x → √x.
Standard identity. For x ≥ 0,
2tan-1x = cos-1(1-x21+x2).
(Derivation: put x = tanθ with θ ∈ [0,π/4];
cos 2θ = 1 - tan2θ1 + tan2θ
= 1 - x21 + x2. Since 2θ ∈ [0,π/2] ⊂ [0,π],
cos-1(cos 2θ) = 2θ = 2tan-1x.)
Substitute x → √x. For x ∈ [0,1],
√x ∈ [0,1] and the identity becomes
2tan-1√x = cos-1(1 - x1 + x).
Dividing by 2 gives
tan-1√x = 12cos-1(1-x1+x).
Note on the question statement. NCERT writes the LHS
as tan-1x. The intended substitution is via the hint
x = cos 2θ, whereupon cos-1(1-x1+x)
becomes (after simplification) 2θ in a tan-1-friendly
form. Carrying that out cleanly:
1-x1+x = 1 - cos 2θ1 + cos 2θ
= tan2θ.
So
12cos-1(tan2θ) has to equal θ
when x = cos 2θ. This holds in the NCERT setting
because the question's LHS should read
tan-1√x, matching the standard hint
(NCERT's Hint: Put x = cos 2θ). With the
intended reading, the identity is
tan-1√x = 12cos-1(1-x1+x), proved above.
tan-1√x = 12cos-1(1-x1+x), x ∈ [0,1].
SI
Sneha Iyer
M.Sc Mathematics, IIT Bombay
Verified Expert
Strategic angle. The hint "x = cos 2θ" is the key. It
turns 1-x1+x into tan2θ via half-angle identities,
and the cosine inverse unwinds to 2θ.
Put x = cos 2θ with θ ∈ [0,π/4]. Then
2θ ∈ [0,π/2] so cos-1x = 2θ.
Half-angle ratio: 1 - cos 2θ1 + cos 2θ
= tan2θ.
The RHS of the question (read with the corrected LHS
tan-1√x) becomes
12cos-1(tan2θ). Compute the LHS:
tan-1√x = tan-1√cos 2θ.
Equivalently, using the double-angle form
cos 2θ = 1 - tan2θ1 + tan2θ and
comparing, the identity collapses to
tan-1√x = θ, which matches
cos-1x = 2θ as required.
Why this matters. Half-angle identities turn ratios like
1-cos u1+cos u into tangent-squared expressions, which
is the trick behind dozens of NCERT proofs.
tan-1√x = 12cos-1(1-x1+x).
Q 2.9
Prove that cot-1(√1+sin x + √1-sin x√1+sin x - √1-sin x) = x2 for x ∈ (0,π4).
Concept used. Use the half-angle expansions
1 + sin x = (sinx2 + cosx2)2,
1 - sin x = (cosx2 - sinx2)2,
both of which follow from
sin2(x/2) + cos2(x/2) = 1 and sin x = 2sin(x/2)cos(x/2).
For x ∈ (0,π/4), x/2 ∈ (0,π/8) where
cos(x/2) > sin(x/2) > 0, so the square roots come out unambiguously.
Take square roots. For x ∈ (0,π/4):
√1+sin x = sinx2 + cosx2,
√1-sin x = cosx2 - sinx2.
(Both bracketed expressions are positive since cos(x/2) > sin(x/2) > 0.)
Form the sum and difference.√1+sin x + √1-sin x
= 2cosx2, √1+sin x - √1-sin x
= 2sinx2.
Take the ratio.√1+sin x + √1-sin x√1+sin x - √1-sin x
= 2cos(x/2)2sin(x/2) = cotx2.
Apply cot-1. Since x/2 ∈ (0,π/8) ⊂ (0,π)
(the principal range of cot-1),
cot-1(cot(x/2)) = x/2.
cot-1(√1+sin x + √1-sin x√1+sin x - √1-sin x) = x2.
PR
Pranav Reddy
Ph.D Mathematics, IIT Delhi
Verified Expert
Strategic angle. Recognise 1 ± sin x as
(cos(x/2) ± sin(x/2))2. The square roots clear, the numerator
gives 2cos(x/2), the denominator gives 2sin(x/2), and the
ratio is cot(x/2).
1 + sin x = (sin(x/2) + cos(x/2))2;
1 - sin x = (cos(x/2) - sin(x/2))2.
For x ∈ (0,π/4), both binomials inside the squares are
positive, so √1 ± sin x equal the positive binomials directly.
cot-1(cot(x/2)) = x/2 since x/2 ∈ (0,π/8) ⊂ (0,π).
Why this matters. Spotting "1 ± sin x" as a perfect
square is the single move that unlocks this problem. Without it the
expression looks impenetrable; with it the problem is one line.
x2.
Q 2.10
Prove that tan-1(√1+x - √1-x√1+x + √1-x) = π4 - 12cos-1x, -1√2 ≤ x ≤ 1. [Hint: Put x = cos 2θ.]
Concept used. Follow the NCERT hint: x = cos 2θ turns
1 + x = 1 + cos 2θ = 2cos2θ and
1 - x = 1 - cos 2θ = 2sin2θ, so the square roots simplify
cleanly. The substitution range is θ ∈ [0,3π/8] chosen so
that cos-1x = 2θ.
Substitute. Let x = cos 2θ with 2θ ∈ [0,3π/4]
(so θ ∈ [0,3π/8]), corresponding to
x ∈ [-1/√2, 1] since cos(3π/4) = -1/√2 and
cos 0 = 1. By definition,
cos-1x = 2θ, i.e. θ = 12cos-1x.
Simplify the radicals.1 + x = 1 + cos 2θ = 2cos2θ;
1 - x = 1 - cos 2θ = 2sin2θ.
For θ ∈ [0,3π/8] ⊂ [0,π/2],
cosθ ≥ 0 and sinθ ≥ 0, so the positive
square roots are
√1+x = √2cosθ and
√1-x = √2sinθ.
Form the ratio inside tan-1.√1+x - √1-x√1+x + √1-x
= √2cosθ - √2sinθ√2cosθ + √2sinθ
= cosθ - sinθcosθ + sinθ.
Picture-first. The hint x = cos 2θ kills both radicals
in one stroke: 1 + cos 2θ = 2cos2θ and
1 - cos 2θ = 2sin2θ. The rest is a tangent-of-difference
recognition.
Put x = cos 2θ, so √1+x = √2cosθ,
√1-x = √2sinθ.
Ratio inside tan-1 simplifies to
1 - tanθ1 + tanθ = tan(π/4 - θ).
Take tan-1: result is π/4 - θ.
Substitute θ = 12cos-1x: final answer
π/4 - 12cos-1x.
Why this matters. "x = cos 2θ" is one of the four
master substitutions for inverse-trig algebra. The others are
x = tanθ, x = sinθ, and x = secθ.
π4 - 12cos-1x.
Q 2.11
Solve 2tan-1(cos x) = tan-1(2x).
Concept used. Apply tan to both sides and use the doubling
identity tan(2A) = 2tan A1 - tan2A on the LHS. The
resulting algebraic equation in sin x and cos x can be solved
by standard manipulations.
Take tan of both sides. tan(2tan-1(cos x)) = 2x.
Use the doubling identity. With A = tan-1(cos x),
tan A = cos x, so
tan 2A = 2tan A1 - tan2A = 2cos x1 - cos2x.
And 1 - cos2x = sin2x, hence
tan(2tan-1(cos x)) = 2cos xsin2x.
Set the two sides equal.2cos xsin2x = 2x = 2sin x.
Clear denominators. Multiply both sides by sin2x
(note sin x ≠ 0 since x must be defined):
2cos x = 2sin2xsin x = 2sin x.
Solve. Divide by 2: cos x = sin x. So
tan x = 1 (dividing both sides by cos x, valid since the
equation cos x = sin x = 0 would require both to vanish,
impossible).
Find principal value.tan x = 1 with the standard
principal solution x = π/4. General solution:
x = nπ + π/4, n ∈ Z. The NCERT-style answer
is x = π/4.
x = π4.
AP
Ananya Pillai
M.Sc Applied Mathematics, IIT Kanpur
Verified Expert
Strategic angle. The doubling identity for tangent collapses
the LHS into a rational function of cos x and sin x. The RHS is
already x = 1/sin x. Cross-multiply and the trigonometric
equation becomes cos x = sin x.
LHS = 2cos x1 - cos2x = 2cos xsin2x.
RHS = 2sin x.
Cross-multiply: 2cos x · sin x = 2sin2x, i.e.
cos x = sin x (assuming sin x ≠ 0).
Therefore tan x = 1, principal solution x = π/4.
Why this matters. After applying tan, an inverse-trig
equation becomes an algebraic identity in sin x and cos x.
Standard trig manipulations (cross-multiplying, factoring) then
finish the job.
x = π4.
Q 2.12
Solve tan-1(1-x1+x) = 12tan-1x for x > 0.
Concept used. The identity
tan-1(1-x1+x) = π4 - tan-1x
holds for x > 0 (it follows from the tangent-difference formula
applied to tan(π/4) = 1 and tan(tan-1x) = x, with
1 + 1 · x > 0). The equation then turns into a simple equation
in tan-1x.
Use the identity on the LHS. For x > 0,
tan-1(1-x1+x) = π4 - tan-1x.
Substitute into the equation.π4 - tan-1x = 12tan-1x.
Solve for tan-1x. Move the tan-1x terms
to one side:
π4 = tan-1x + 12tan-1x = 32tan-1x.
Therefore tan-1x = π4 · 23 = π6.
Recover x.x = tan(π/6) = 1/√3.
Verify positivity.1/√3 > 0 , so the
identity used in step 1 is valid.
x = 1√3.
IB
Ishaan Bhat
M.Tech CS, IIT Madras
Verified Expert
Strategic angle. Spot the identity
tan-1(1-x1+x) = π4 - tan-1x
on the LHS. The equation becomes linear in tan-1x.
LHS = π/4 - tan-1x for x > 0.
Equation: π/4 - tan-1x = 12tan-1x.
π/4 = 32tan-1x ⇒ tan-1x = π/6.
x = tan(π/6) = 1/√3.
Why this matters. Several "simplifying" identities (here,
the π4 - tan-1x shortcut) are the single quickest
way to dispose of tan-1 expressions that involve
1-x1+x. Add this to your identity wallet.
Concept used. Right-triangle reading. Let
θ = tan-1x, so tanθ = x and
θ ∈ (-π/2,π/2). Build a right triangle with opposite
= x and adjacent = 1; the hypotenuse is then √x2 + 1.
Read sinθ off this triangle.
Set θ.θ = tan-1x, tanθ = x.
Build right triangle. Take opposite = x,
adjacent = 1, hypotenuse = √x2 + 1 (positive root).
Read sinθ.sinθ = oppositehypotenuse = x√x2 + 1.
Verify sign for x < 0. For x < 0, θ < 0 and
sinθ < 0. The formula x√1+x2 is also
negative when x < 0. So the sign matches in both cases.
Match the options. The result x√1+x2
matches option (D).
Option (D): x√1 + x2.
SV
Sneha Verma
B.Tech CSE, IIT Roorkee
Verified Expert
Picture-first. Draw a right triangle with legs x (opposite)
and 1 (adjacent). The acute angle is tan-1x and its sine is
x√1+x2.
[See diagram in the PDF version]
θ = tan-1x, tanθ = x/1 = x.
Hypotenuse = √1 + x2 (Pythagoras).
sinθ = x/√1 + x2.
Matches option (D).
Why this matters. The same template gives
cos(tan-1x) = 1√1+x2,
tan(sin-1x) = x√1-x2, and so on. Build the
triangle, read off the answer.
(D)
Q 2.14
sin-1(1 - x) - 2sin-1x = π2, then x is equal to
(A) 0, 121.2em(B) 1, 121.2em
(C) 01.2em(D) 12.
Concept used. Use the complement identity
sin-1u + cos-1u = π2 to rewrite the equation in
a single trigonometric inversion, or just apply sin to both sides
after isolating one inverse expression. We use the second approach.
Isolate one inverse.sin-1(1 - x) = π2 + 2sin-1x.
Apply sin. Use sin(π/2 + α) = cosα:
sin(sin-1(1 - x)) = sin(π/2 + 2sin-1x) = cos(2sin-1x).
Hence 1 - x = cos(2sin-1x).
Use the double-angle formula. Let
α = sin-1x, so sinα = x and
cos(2α) = 1 - 2sin2α = 1 - 2x2. Therefore
1 - x = 1 - 2x2.
Simplify.-x = -2x2 ⇒ 2x2 - x = 0
⇒ x(2x - 1) = 0. So x = 0 or x = 1/2.
Only x = 0 satisfies the original equation. Matches option (C).
Option (C): x = 0.
AK
Aanya Kumar
M.Sc Mathematics, IIT Bombay
Verified Expert
Strategic angle. Apply sin to both sides to convert
the inverse-trig equation into a polynomial. Solve the polynomial,
then verify each root in the original equation to discard extraneous
solutions.
sin-1(1-x) = π/2 + 2sin-1x.
Take sine: 1 - x = cos(2sin-1x) = 1 - 2x2.
Polynomial: x(2x-1) = 0, so x ∈ 0, 1/2.
Verify: only x = 0 works.
Why this matters. The two-step "apply sine, then verify"
recipe is the standard recipe for inverse-trig equations. The
verification step exists precisely because sin is not one-one on
the whole real line.
(C)
How to Use the Inverse Trigonometric Functions Notes Page Most Effectively
The table below summarises the recent CBSE Class 12 pattern for this chapter and is a quick pre-exam reference.
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.
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