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Exercise - 2.2

Inverse Trigonometry

**Question-1 :-** Prove that 3sin^{-1} x = sin^{-1} (3x - 4x³), x ∈ [-1/2, 1/2].

Let x = sin θ and θ = sin^{-1}x. R.H.S sin^{-1}(3x - 4x³) = sin^{-1}(3sin θ - 4sin³ θ) = sin^{-1}(sin 3θ) = 3θ = 3sin^{-1}x

**Question-2 :-** Prove that 3cos^{-1} x = cos^{-1} (4x³ - 3x), x ∈ [1/2, 1].

Let x = cos θ and θ = cos^{-1}x. R.H.S cos^{-1}(4x³ - 3x) = cos^{-1}(4cos³ θ - 3cos θ) = cos^{-1}(cos 3θ) = 3θ = 3cos^{-1}x

**Question-3 :-** Prove that
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**Question-4 :-** Prove that
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**Question-5 :-** Write the function in simplest form :
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**Question-6 :-** Write the function in simplest form :
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**Question-7 :-** Write the function in simplest form :
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**Question-8 :-** Write the function in simplest form :
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**Question-9 :-** Write the function in simplest form :
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**Question-10 :-** Write the function in simplest form :
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**Question-11 :-** Find the value of.

**Question-12 :-** Find the value of cot (tan^{-1}a + cot^{-1}a).

cot(tan^{-1}a + cot^{-1}a) = cot(π/2) ∴ [tan^{-1}a + cot^{-1}a = π/2] = 0

**Question-13 :-** Find the value of.

**Question-14 :-** Find the value of x, if sin (sin^{-1} 1/5 + cos^{-1} x) = 1.

sin (sin^{-1}1/5 + cos^{-1}x) = 1 sin (sin^{-1}1/5 + cos^{-1}x) = sin π/2 sin^{-1}1/5 + cos^{-1}x = sin^{-1}(sin π/2) sin^{-1}1/5 + cos^{-1}x = π/2 ∴ [sin^{-1}x + cos^{-1}x = π/2] sin^{-1}1/5 = π/2 - cos^{-1}x sin^{-1}1/5 = sin^{-1}x 1/5 = x or x = 1/5

**Question-15 :-** Find the value of.

**Question-16 :-** Find the value of sin^{-1}(sin 2π/3).

sin^{-1}(sin 2π/3) Range of sin^{-1}is [-π/2, π/2], so 2π/3 is not in Range. Then we break 2π/3. = sin^{-1}[sin(π - π/3)] = sin^{-1}[sin π/3] = π/3

**Question-17 :-** Find the value of tan^{-1}(tan 3π/4).

tan^{-1}(tan 3π/4) Range of tan^{-1}is [-π/2, π/2], so 3π/4 is not in Range. Then we break 3π/4. = tan^{-1}[tan(π - π/4)] = tan^{-1}[-tan π/4] = tan^{-1}[tan (-π/4)] = -π/4

**Question-18 :-** Find the value of tan(sin^{-1} 3/5 + cot^{-1} 3/2).

tan(sin^{-1}3/5 + cot^{-1}3/2) Firstly, Let sin^{-1}3/5 = θ₁ sin θ₁ = 3/5 cos θ₁ = √1 - sin²θ₁ cos θ₁ = √1 - (3/5)² cos θ₁ = √1 - 9/25 cos θ₁ = √16/25 cos θ₁ = 4/5 Then, tan θ₁ = sin θ₁/cos θ₁ = 3/5 x 5/4 = 3/4 tan θ₁ = 3/4 θ₁ = tan^{-1}3/4 Secondly, Let cot^{-1}3/2 = θ₂ cot θ₂ = 3/2 tan θ₂ = 1/cot θ₂ tan θ₂ = 2/3 θ₂ = tan^{-1}2/3 Now, Accordin to question tan(θ₁ + θ₂) = tan(tan^{-1}3/4 + tan^{-1}2/3) = tan[tan^{-1}(3/4 + 2/3) ÷ (1 - 3/4 x 2/3)] = tan[tan^{-1}(9 + 8)/12 ÷ (1 - 1/2)] = tan[tan^{-1}(17/12 ÷ 1/2)] = tan[tan^{-1}(17/12 x 2/1)] = tan[tan^{-1}(17/6)] = 17/6

**Question-19 :-** cos^{-1}(cos 7π/6) is equal to

(A) 7π/6 (B) 5π/6 (C) π/3 (D) π/6

cos^{-1}(cos 7π/6) Range of cos^{-1}is [0, π], so 7π/6 is not in Range. Then we break 7π/6. = cos^{-1}[cos(π + π/6)] = cos^{-1}[-cos π/6] = cos^{-1}[cos (-π/6)] {Range = [0, π]} = cos^{-1}[cos (π - π/6)] = cos^{-1}[cos (5π/6)] = 5π/6 Option is (B)

**Question-20 :-** sin [π/3 + sin^{-1}(-1/2)] is equal to

(A) 1/2 (B) 1/3 (C) 1/4 (D) 1

sin [π/3 + sin^{-1}(-1/2)] = sin [π/3 + sin^{-1}(-sin π/6)] = sin [π/3 + sin^{-1}(sin (-π/6))] = sin [π/3 - π/6)] = sin [(2π - π)/6] = sin [π/6] = 1/2

**Question-21 :-** tan^{-1}(√3) - cot^{-1}(-√3) is equal to

(A) π (B) -π/2 (C) 0 (D) 2√3

tan^{-1}(√3) - cot^{-1}(-√3) Let tan^{-1}(√3) = θ₁ tan θ₁ = √3 {Range = [-π/2, π/2]} tan θ₁ = tan π/3 θ₁ = π/3 Let cot^{-1}(-√3) = θ₂ cot θ₂ = -√3 cot θ₂ = -cot π/6 cot θ₂ = cot (-π/6) {Range = [0, π]} cot θ₂ = cot (π - π/6) cot θ₂ = cot (5π/6) θ₂ = 5π/6 Now, θ₁ - θ₂ = π/3 - 5π/6 = (2π - 5π)/6 = -3π/6 = -π/2 Option (B) is correct.

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