Find x , if sin - 1 1 - x = π 2 + 2 sin - 1 x .

Given that sin - 1 1 - x = π 2 + 2 sin - 1 x ⇒ sin - 1 1 - x - 2 sin - 1 x = π 2 Let y = sin - 1 x ⇒ sin - 1 1 - sin y - 2 y = π 2 ⇒ sin - 1 1 - sin y = 2 y + π 2 ⇒ 1 - sin y = sin 2 y + π 2 = cos 2 y = 1 - 2 sin 2 y ⇒ 2 sin 2 y - sin y = 0 ⇒ 2 x 2 - x = 0 ⇒ x 2 x - 1 = 0 ⇒ x = 0 , x = 1 2 Since, x = 1 2 does not satisfy the given equation. So, x = 0 is the solution of given equation .

Find the value of sin 1 2 cot - 1 - 3 4 .

Let y = sin 1 2 cot - 1 - 3 4 Consider t = cot - 1 - 3 4 ⇒ cot t = - 3 4 , where π 2 < t < π ⇒ cosec 2 t = cot 2 t + 1 = 9 16 + 1 = 25 16 ⇒ cosec t = 5 4 ⇒ sin t = 4 5 Since, cos 2 t = 1 - sin 2 t = 1 - 16 25 = 9 25 ⇒ cos t = ± 3 5 Since π 2 < t < π ⇒ cos t = - 3 5 Now, y = sin t 2 = 1 2 1 - cos t ⇒ y = 1 + 3 5 2 = 2 2 5 × 2 = 2 5 .

Simplify sin - 1 x + sin - 1 1 - x 2 , where - 1 ≤ x ≤ 1 .

Consider y = sin - 1 x + sin - 1 1 - x 2 , where - 1 ≤ x ≤ 1 Let x = sin θ ⇒ 1 - x 2 = 1 - sin 2 θ = cos 2 θ ⇒ y = sin - 1 sin θ + sin - 1 cos 2 θ ⇒ y = θ + sin - 1 cos θ = θ + sin - 1 sin π 2 - θ ⇒ y = θ + π 2 - θ = π 2 .

Find the value of tan - 1 1 2 tan 2 A + tan - 1 cot A + tan - 1 cot 3 A , 0 < A < π 4 .

Let y = tan - 1 1 2 tan 2 A + tan - 1 cot A + tan - 1 cot 3 A Since 0 < A < π 4 ⇒ cot A > 1 , cot 3 A > 1 ⇒ cot A cot 3 A > 1 We know that tan - 1 x + tan - 1 y = π + tan - 1 x + y 1 - x y , x > 0 , y > 0 , x y > 1 ⇒ y = tan - 1 1 2 × 2 tan A 1 - tan 2 A + π + tan - 1 cot A + cot 3 A 1 - cot 4 A = tan - 1 tan A 1 - tan 2 A + π + tan - 1 cot A 1 + cot 2 A 1 + cot 2 A 1 - cot 2 A = tan - 1 tan A 1 - tan 2 A + π + tan - 1 tan A tan 2 A - 1 = tan - 1 tan A 1 - tan 2 A + π - tan - 1 tan A 1 - tan 2 A = π .

Find the value of tan - 1 2 cos 2 sin - 1 1 2 + cot - 1 2 sin 2 cos - 1 1 2 .

Let y = tan - 1 2 cos 2 sin - 1 1 2 + cot - 1 2 sin 2 cos - 1 1 2 = tan - 1 2 cos 2 × π 6 + cot - 1 2 sin 2 × π 3 = tan - 1 2 cos π 3 + cot - 1 2 sin 2 π 3 = tan - 1 2 × 1 2 + cot - 1 2 × 3 2 = tan - 1 1 + cot - 1 3 = π 4 + π 6 = 5 π 12

Prove that 2 tan - 1 1 3 + sin - 1 4 5 = π 2 .

L . H . S . = 2 tan - 1 1 3 + sin - 1 4 5 We know that 2 tan - 1 x = cos - 1 1 - x 2 1 + x 2 , if - 1 ≤ x ≤ 1 As 1 3 ∈ - 1 , 1 2 tan - 1 1 3 = cos - 1 1 - 1 3 2 1 + 1 3 2 = cos - 1 8 10 = cos - 1 4 5 ∴ 2 tan - 1 1 3 + sin - 1 4 5 = cos - 1 4 5 + sin - 1 4 5 = π 2 = R . H . S .

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Gamma Question Bank for Engineering Mathematics>Class XII>Chapter 22 - Inverse Trigonometric Functions>Assignment

Embibe Experts Solutions for Exercise 4: Assignment

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Exercise 4: Assignment

Attempt the free practice questions from Exercise 4: Assignment with hints and solutions to strengthen your understanding. Gamma Question Bank for Engineering Mathematics solutions are prepared by Experienced Embibe Experts.