Question-1 :- Prove that 3sin-1 x = sin-1 (3x - 4x³), x ∈ [-1/2, 1/2].
Solution :-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].
Solution :-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-5 :- Write the function in simplest form :
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Question-6 :- Write the function in simplest form :
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Question-11 :- Find the value of.
Question-12 :- Find the value of cot (tan-1a + cot-1a).
Solution :-cot(tan-1a + cot-1a) = cot(π/2) ∴ [tan-1a + cot-1a = 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.
Solution :-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-1x + cos-1x = 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).
Solution :-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).
Solution :-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).
Solution :-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.