Ncert Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions Misc

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.

1. Check. 13π6 = 2π + π6, so 13π6 > 2π > π, well outside the principal range [0,π].
2. Reduce modulo 2π. Use cos(u - 2π) = cos u: cos(13π6) = cos(13π6 - 2π) = cos(13π - 12π6) = cos(π6).
3. Check. π6 ∈ [0,π] .
4. Apply cos^-1. cos^-1(cos13π6) = cos^-1(cosπ6) = π6.

cos^-1(cos13π6) = π6.

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.

1. Check. 7π6 > π2 = 3π6, so outside (-π/2,π/2).
2. Subtract π. tan(u - π) = tan u gives tan(7π6) = tan(7π6 - π) = tan(7π - 6π6) = tan(π6).
3. Check. π6 ∈ (-π/2, π/2) .
4. Apply tan^-1. tan^-1(tan7π6) = tan^-1(tanπ6) = π6.

tan^-1(tan7π6) = π6.

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.

1. Set θ. Let θ = sin^-1(3/5), so sinθ = 3/5 and θ ∈ (0,π/2) since 3/5 > 0.
2. 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.
3. Compute tan 2θ. Apply the double-angle formula: tan 2θ = 2tanθ1 - tan²θ = 2 · 341 - 916 = 32716.
4. Simplify. 3/27/16 = 32 · 167 = 4814 = 247.
5. 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.
6. Conclude. 2θ = tan^-1(24/7), i.e. 2sin^-1(3/5) = tan^-1(24/7).

2sin^-135 = tan^-1247.

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.

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.
2. Build triangle for B. sin B = 3/5 gives cos B = 4/5 and tan B = 3/4 (the standard 3,4,5 triangle).
3. Apply tangent sum. tan(A + B) = tan A + tan B1 - tan A tan B = 815 + 341 - 815 · 34.
4. Simplify numerator. Common denominator 60: 815 = 3260, 34 = 4560, so 815 + 34 = 32 + 4560 = 7760.
5. Simplify denominator. 815 · 34 = 2460 = 25, so 1 - 25 = 35.
6. Divide. tan(A+B) = 77/603/5 = 7760 · 53 = 7736.
7. 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.

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.

1. For A: cos A = 4/5, hence by sin² A + cos² A = 1, sin A = √1 - (4/5)² = √1 - 16/25 = √9/25 = 3/5 (positive since A ∈ [0,π/2]).
2. For B: cos B = 12/13, hence sin B = √1 - 144/169 = √25/169 = 5/13.
3. Cosine sum formula. cos(A+B) = cos A cos B - sin A sin B = 45 · 1213 - 35 · 513.
4. Compute the two products. 45 · 1213 = 4865; 35 · 513 = 1565.
5. Subtract. cos(A+B) = 4865 - 1565 = 3365.
6. 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.

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.

1. For A: cos A = 12/13, so sin A = √1 - 144/169 = 5/13 (positive in [0,π/2]).
2. For B: sin B = 3/5, so cos B = √1 - 9/25 = 4/5 (positive in [0,π/2]).
3. Sine sum formula. sin(A+B) = sin A cos B + cos A sin B = 513 · 45 + 1213 · 35.
4. Compute the two products. 513 · 45 = 2065; 1213 · 35 = 3665.
5. Add. sin(A+B) = 2065 + 3665 = 5665.
6. 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.

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.

1. For A: sin A = 5/13, cos A = 12/13 (triple (5,12,13)), so tan A = 5/12.
2. For B: cos B = 3/5, sin B = 4/5 (triple (3,4,5)), so tan B = 4/3.
3. Apply tangent sum. tan(A+B) = tan A + tan B1 - tan A tan B = 512 + 431 - 512 · 43.
4. Simplify numerator. Common denominator 12: 512 + 43 = 512 + 1612 = 2112.
5. Simplify denominator. 512 · 43 = 2036 = 59, so 1 - 59 = 49.
6. Divide. tan(A+B) = 21/124/9 = 2112 · 94 = 18948 = 6316 (dividing numerator and denominator by 3).
7. 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.

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.

1. 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].
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.
3. 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.
4. Apply cos^-1 on the RHS. The expression becomes cos^-1(tan²φ), which is not obviously clean. So we rethink: use instead x = tan² t. 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.
5. 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.)
6. 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).
7. 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].

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.

1. 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.)
2. Form the sum and difference. √1+sin x + √1-sin x = 2cosx2, √1+sin x - √1-sin x = 2sinx2.
3. Take the ratio. √1+sin x + √1-sin x√1+sin x - √1-sin x = 2cos(x/2)2sin(x/2) = cotx2.
4. 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.

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.

1. 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.
2. 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.
3. Form the ratio inside tan^-1. √1+x - √1-x√1+x + √1-x = √2cosθ - √2sinθ√2cosθ + √2sinθ = cosθ - sinθcosθ + sinθ.
4. Divide by cosθ. cosθ - sinθcosθ + sinθ = 1 - tanθ1 + tanθ = tan(π/4) - tanθ1 + tan(π/4)tanθ = tan(π4 - θ).
5. Apply tan^-1. For θ ∈ [0,3π/8], π4 - θ ∈ [π4 - 3π8, π4] = [-π8, π4] ⊂ (-π2,π2). Hence tan^-1(tan(π4 - θ)) = π4 - θ.
6. Back-substitute. θ = 12cos^-1x, so tan^-1(√1+x - √1-x√1+x + √1-x) = π4 - 12cos^-1x.

tan^-1(√1+x - √1-x√1+x + √1-x) = π4 - 12cos^-1x.

Concept used. Apply tan to both sides and use the doubling identity tan(2A) = 2tan A1 - tan² A on the LHS. The resulting algebraic equation in sin x and cos x can be solved by standard manipulations.

1. Take tan of both sides. tan(2tan^-1(cos x)) = 2x.
2. Use the doubling identity. With A = tan^-1(cos x), tan A = cos x, so tan 2A = 2tan A1 - tan² A = 2cos x1 - cos² x. And 1 - cos² x = sin² x, hence tan(2tan^-1(cos x)) = 2cos xsin² x.
3. Set the two sides equal. 2cos xsin² x = 2x = 2sin x.
4. Clear denominators. Multiply both sides by sin² x (note sin x ≠ 0 since x must be defined): 2cos x = 2sin² xsin x = 2sin x.
5. 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).
6. 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.

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.

1. Use the identity on the LHS. For x > 0, tan^-1(1-x1+x) = π4 - tan^-1x.
2. Substitute into the equation. π4 - tan^-1x = 12tan^-1x.
3. 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.
4. Recover x. x = tan(π/6) = 1/√3.
5. Verify positivity. 1/√3 > 0 , so the identity used in step 1 is valid.

x = 1√3.

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.

1. Set θ. θ = tan^-1x, tanθ = x.
2. Build right triangle. Take opposite = x, adjacent = 1, hypotenuse = √x² + 1 (positive root).
3. Read sinθ. sinθ = oppositehypotenuse = x√x² + 1.
4. 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.
5. Match the options. The result x√1+x² matches option (D).

Option (D): x√1 + x².

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.

1. Isolate one inverse. sin^-1(1 - x) = π2 + 2sin^-1x.
2. 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).
3. Use the double-angle formula. Let α = sin^-1x, so sinα = x and cos(2α) = 1 - 2sin²α = 1 - 2x². Therefore 1 - x = 1 - 2x².
4. Simplify. -x = -2x² ⇒ 2x² - x = 0 ⇒ x(2x - 1) = 0. So x = 0 or x = 1/2.
5. Check each candidate in the original equation.
   • x = 0: LHS = sin^-1(1) - 2sin^-1(0) = π/2 - 0 = π/2 .
   • x = 1/2: LHS = sin^-1(1/2) - 2sin^-1(1/2) = -sin^-1(1/2) = -π/6 ≠ π/2. Reject.
   Only x = 0 satisfies the original equation. Matches option (C).

Option (C): x = 0.