Ncert Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2 2

Concept used. The triple-angle formula for sine states sin 3θ = 3sinθ - 4sin³θ. This is derived from sin 3θ = sin(2θ+θ) = sin 2θ cosθ + cos 2θ sinθ and the double-angle formulas. Combined with the principal-range identity sin^-1(sin u) = u provided u ∈ [-π2,π2], it gives the inverse-trig identity asked here. The condition x ∈ [-12,12] is precisely what guarantees 3θ ∈ [-π2,π2] in the step where we apply this principal-range identity.

1. Substitute. Let x = sinθ. By the definition of sin^-1, this is equivalent to θ = sin^-1x ∈ [-π2,π2].
2. Restrict θ. Since x ∈ [-12,12], sinθ ∈ [-12,12] which forces θ ∈ [-π6,π6] (because sin^-1 is increasing and sin^-1(±12) = ±π6).
3. Hence 3θ stays in the principal range. From θ ∈ [-π6,π6], multiply by 3 to get 3θ ∈ [-π2,π2]. This is the key bound that lets us apply sin^-1(sin·) later.
4. Apply the triple-angle formula. Compute 3x - 4x³: 3x - 4x³ = 3sinθ - 4sin³θ = sin 3θ.
5. Apply sin^-1 on both sides. Take sin^-1: sin^-1(3x - 4x³) = sin^-1(sin 3θ) = 3θ, where the last equality uses 3θ ∈ [-π2,π2].
6. Recover the right-hand side. Since θ = sin^-1x, 3θ = 3sin^-1x, so sin^-1(3x - 4x³) = 3sin^-1x.

3sin^-1x = sin^-1(3x - 4x³) for x ∈ [-12,12].

Concept used. The triple-angle formula for cosine is cos 3θ = 4cos³θ - 3cosθ. This is derived from cos 3θ = cos(2θ+θ) via the double-angle and addition formulas. We combine it with the principal-range identity cos^-1(cos u) = u for u ∈ [0,π]. The bound x ∈ [12,1] guarantees 3θ stays inside [0,π].

1. Substitute. Let x = cosθ. By the definition of cos^-1, equivalently θ = cos^-1x ∈ [0,π].
2. Restrict θ. Since x ∈ [12,1] and cos^-1 is decreasing, cos^-1(1) = 0 and cos^-1(12) = π3, hence θ ∈ [0,π3].
3. So 3θ stays in [0,π]. Multiply θ ∈ [0,π3] by 3 to get 3θ ∈ [0,π].
4. Apply the triple-angle formula. Compute 4x³ - 3x = 4cos³θ - 3cosθ = cos 3θ.
5. Apply cos^-1. Since 3θ ∈ [0,π], cos^-1(4x³ - 3x) = cos^-1(cos 3θ) = 3θ = 3cos^-1x.

3cos^-1x = cos^-1(4x³ - 3x) for x ∈ [12,1].

Concept used. The substitution x = tanθ is the canonical way to simplify any inverse-tangent expression containing √1+x², because that radical becomes secθ and trig identities collapse the algebra. We also need the half-angle identities 1 - cosθ = 2sin²(θ/2) and sinθ = 2sin(θ/2)cos(θ/2).

1. Substitute. Let x = tanθ with θ ∈ (-π2,π2). Then √1+x² = √1+tan²θ = √sec²θ = secθ, with the positive root because secθ > 0 on the principal range of tan^-1.
2. Rewrite the argument. Substitute x = tanθ and √1+x² = secθ: √1+x²-1x = secθ - 1tanθ = 1cosθ - 1sinθcosθ = 1 - cosθsinθ. (Multiplied top and bottom by cosθ.)
3. Apply half-angle identities. 1 - cosθsinθ = 2sin²(θ/2)2sin(θ/2)cos(θ/2) = tan(θ2).
4. Apply tan^-1. The half-angle θ2 lies in (-π4,π4), well inside the principal range, so tan^-1(tan(θ/2)) = θ/2.
5. Back-substitute θ = tan^-1x. tan^-1(√1+x²-1x) = 12tan^-1x.

tan^-1(√1+x² - 1x) = 12tan^-1x.

Concept used. The half-angle identities 1 - cos x = 2sin²(x/2) and 1 + cos x = 2cos²(x/2). Their ratio reduces to tan²(x/2), and the principal-range identity tan^-1(tan u) = u for u ∈ (-π/2,π/2) finishes the work.

1. Rewrite the radicand. Substitute the half-angle identities: 1-cos x1+cos x = 2sin²(x/2)2cos²(x/2) = tan²(x/2).
2. Take the square root. Since 0 < x < π, we have 0 < x/2 < π/2, so tan(x/2) > 0. Hence √tan²(x/2) = tan(x/2).
3. Apply tan^-1. Because x/2 ∈ (0,π/2) ⊂ (-π/2,π/2), tan^-1(tan(x/2)) = x/2.

tan^-1(√1-cos x1+cos x) = x2.

Concept used. Divide numerator and denominator by cos x to expose tan x, then apply the tangent subtraction formula tan(A - B) = tan A - tan B1 + tan A tan B with A = π4 (so tan A = 1). The principal-range identity tan^-1(tan u) = u for u ∈ (-π/2,π/2) finishes the work.

1. Divide top and bottom by cos x. cos x - sin xcos x + sin x = 1 - tan x1 + tan x.
2. Recognise the tangent of a difference. Write 1 = tan(π/4): 1 - tan x1 + tan x = tan(π/4) - tan x1 + tan(π/4)tan x = tan(π4 - x).
3. Check principal-range membership. As x runs through (-π4,3π4), u = π4 - x runs through (-π2,π2): x = -π4 gives u = π2 and x = 3π4 gives u = -π2, with the open interval inside.
4. Apply tan^-1. tan^-1(tan(π/4 - x)) = π/4 - x.

tan^-1(cos x - sin xcos x + sin x) = π4 - x.

Concept used. The trigonometric substitution x = asinθ converts √a² - x² to acosθ, after which the argument of tan^-1 reduces to tanθ.

1. Substitute. Let x = asinθ with θ ∈ [-π2,π2]. Then sinθ = x/a and θ = sin^-1(x/a).
2. Simplify the radical. Since θ ∈ [-π2,π2], cosθ ≥ 0, so √a² - x² = √a² - a²sin²θ = a√1 - sin²θ = acosθ.
3. Rewrite the argument. x√a² - x² = asinθacosθ = tanθ.
4. Apply tan^-1. The condition |x| < a gives |sinθ| < 1, hence θ ∈ (-π2,π2), exactly the principal range of tan^-1: tan^-1(tanθ) = θ = sin^-1(x/a).

tan^-1(x√a² - x²) = sin^-1(xa).

Concept used. The substitution x = atanθ turns the given rational expression into the tangent of 3θ, via the triple-angle formula tan 3θ = 3tanθ - tan³θ1 - 3tan²θ. The bounds on x are crafted so that 3θ stays inside the principal range of tan^-1.

1. Substitute. Let x = atanθ with θ = tan^-1(x/a). The given range -a√3 < x < a√3 gives |tanθ| < 1√3, hence |θ| < π6 and |3θ| < π2.
2. Compute the numerator. With x = atanθ, 3a² x - x³ = 3a²(atanθ) - (atanθ)³ = a³(3tanθ - tan³θ).
3. Compute the denominator. a³ - 3a x² = a³ - 3a(atanθ)² = a³ - 3a³tan²θ = a³(1 - 3tan²θ).
4. Form the ratio. Cancel the common factor a³: 3a² x - x³a³ - 3 a x² = 3tanθ - tan³θ1 - 3tan²θ = tan 3θ.
5. Apply tan^-1. Since |3θ| < π2, tan^-1(3a² x - x³a³ - 3a x²) = tan^-1(tan 3θ) = 3θ = 3tan^-1(x/a).

tan^-1(3a² x - x³a³ - 3 a x²) = 3tan^-1(xa).

Concept used. Work from the innermost inverse outwards. First evaluate sin^-1(1/2) to a principal angle, then take its double-angle cosine, then multiply by 2, then take tan^-1.

1. Innermost inverse. sin^-1(1/2) = π/6 (since sin(π/6) = 1/2 and π/6 ∈ [-π/2,π/2]).
2. Double it. 2sin^-1(1/2) = 2 · π/6 = π/3.
3. Take the cosine. cos(π/3) = 1/2.
4. Multiply by 2. 2cos(2sin^-1(1/2)) = 2 · 1/2 = 1.
5. Outermost inverse. tan^-1(1). Reference: tan(π/4) = 1 and π/4 ∈ (-π/2, π/2), so tan^-1(1) = π/4.

tan^-1(2cos(2sin^-112)) = π4.

Concept used. Two important inverse-trig identities. For |x| ≤ 1, sin^-1(2x1+x²) = 2tan^-1x. For y ≥ 0, cos^-1(1-y²1+y²) = 2tan^-1y. Both follow by setting x = tan A or y = tan B and using the double-angle formulas sin 2A = 2tan A1+tan² A and cos 2B = 1-tan² B1+tan² B. We also use tan(A + B) = tan A + tan B1 - tan A tan B.

1. Simplify the two inverse terms. sin^-1(2x1+x²) = 2tan^-1x and cos^-1(1-y²1+y²) = 2tan^-1y.
2. Multiply each by 12. 12sin^-1(2x1+x²) = tan^-1x, 12cos^-1(1-y²1+y²) = tan^-1y.
3. Add inside the outer tan. The expression becomes tan(tan^-1x + tan^-1y).
4. Apply the tangent sum formula. Let A = tan^-1x, B = tan^-1y, so tan A = x, tan B = y. Then tan(A + B) = tan A + tan B1 - tan A tan B = x + y1 - xy. The condition xy < 1 keeps the denominator positive and non-zero.

tan(12sin^-12x1+x² + 12cos^-11-y²1+y²) = x+y1-xy.

Concept used. The identity sin^-1(sin u) = u holds only when u ∈ [-π2,π2]. For u outside the principal range we use sin(π - u) = sin u to find an equivalent angle inside the principal range.

1. Check the inner angle. Compare 2π3 = 4π6 with π2 = 3π6: 2π3 > π2. Not in the principal range.
2. Find an equivalent angle. Apply sin(π - u) = sin u: sin(2π/3) = sin(π - 2π/3) = sin(π/3).
3. Check. π/3 ∈ [-π/2, π/2] .
4. Apply sin^-1. sin^-1(sin(2π/3)) = sin^-1(sin(π/3)) = π/3.

sin^-1(sin2π3) = π3.

Concept used. tan^-1(tan u) = u only when u ∈ (-π/2,π/2). For u outside this open interval, use the periodicity tan(u - π) = tan u to shift u into the principal range.

1. Check. Compare 3π/4 with π/2 = 2π/4: 3π/4 > π/2, so 3π/4 is not in the principal range.
2. Shift by π. Apply tan(u - π) = tan u: tan(3π/4) = tan(3π/4 - π) = tan(-π/4).
3. Check. -π/2 < -π/4 < π/2 .
4. Apply tan^-1. tan^-1(tan(3π/4)) = tan^-1(tan(-π/4)) = -π/4.

tan^-1(tan3π4) = -π4.

Concept used. Right-triangle reading of sin^-1 and cot^-1. If sin^-1(3/5) = A, then in a right triangle with opposite 3 and hypotenuse 5, the adjacent side is √5² - 3² = 4, giving tan A = 3/4. If cot^-1(3/2) = B, then cot B = 3/2, so tan B = 2/3. The tangent sum formula tan(A + B) = tan A + tan B1 - tan A tan B finishes the work.

1. Find tan A. From A = sin^-1(3/5): sin A = 3/5, cos A = 4/5 (positive in [0,π/2)), so tan A = sin Acos A = 3/54/5 = 34.
2. Find tan B. From B = cot^-1(3/2): cot B = 3/2, hence tan B = 2/3.
3. Apply tangent sum. tan(A+B) = tan A + tan B1 - tan A tan B = 34 + 231 - 34 · 23.
4. Simplify the numerator. Common denominator 12: 34 + 23 = 912 + 812 = 1712.
5. Simplify the denominator. 1 - 34 · 23 = 1 - 612 = 1 - 12 = 12.
6. Divide. tan(A+B) = 17/121/2 = 1712 · 2 = 176.

tan(sin^-135 + cot^-132) = 176.

Concept used. cos^-1(cos u) = u only when u ∈ [0,π]. For u outside this range, use cos(2π - u) = cos u to find an equivalent angle in [0,π].

1. Check. Is 7π/6 in [0,π]? Compare 7π/6 > π = 6π/6. No.
2. Find equivalent angle. Apply cos(2π - u) = cos u: cos(7π/6) = cos(2π - 7π/6) = cos((12π - 7π)/6) = cos(5π/6).
3. Check. 0 ≤ 5π/6 ≤ π .
4. Apply cos^-1. cos^-1(cos(7π/6)) = cos^-1(cos(5π/6)) = 5π/6. Matches option (B).

Option (B): 5π6.

Concept used. Evaluate the inverse first to convert the whole expression into a single sine evaluation.

1. Inner inverse. sin^-1(-1/2) = -sin^-1(1/2) = -π/6.
2. Substitute. Expression becomes sin(π/3 - (-π/6)) = sin(π/3 + π/6).
3. Add fractions. Common denominator 6: π/3 + π/6 = 2π/6 + π/6 = 3π/6 = π/2.
4. Evaluate sin. sin(π/2) = 1.

Option (D): 1.

Concept used. The principal ranges: tan^-1R→(-π/2,π/2) and cot^-1R→(0,π). For a negative argument, cot^-1(-x) = π - cot^-1(x), x ∈ R, following from cot(π - θ) = -cotθ.

1. Compute tan^-1√3. tan(π/3) = √3 and π/3 ∈ (-π/2,π/2), so tan^-1√3 = π/3.
2. Compute cot^-1(√3). cot(π/6) = √3 and π/6 ∈ (0,π), so cot^-1√3 = π/6.
3. Compute cot^-1(-√3). cot^-1(-√3) = π - cot^-1√3 = π - π/6 = 5π/6.
4. Subtract. tan^-1√3 - cot^-1(-√3) = π/3 - 5π/6. Common denominator 6: π/3 = 2π/6, so 2π/6 - 5π/6 = -3π/6 = -π/2.

Option (B): -π2.