Find the value of sin - 1 1 2 - 2 sin - 1 1 2 .

Given that, sin - 1 1 2 - 2 sin - 1 1 2 We know that, 2 Sin - 1 x = Sin - 1 2 x 1 - x 2 . Therefore, sin - 1 1 2 - 2 sin - 1 1 2 = sin - 1 1 2 - sin - 1 2 × 1 2 1 - 1 2 2 On simplifying, we get = sin - 1 1 2 - sin - 1 ( 1 ) = π 6 - π 2 = - π 3

Find the value of sin - 1 cos sin - 1 3 2 .

We know that, cos sin - 1 3 2 = cos ( π / 3 ) Therefore, sin - 1 cos sin - 1 3 2 = sin - 1 cos π 3 = sin - 1 3 2 = π 6

If sin - 1 x 2 + sin - 1 y 2 + sin - 1 z 2 = 3 4 π 2 , find the value of x 2 + y 2 + z 2

We know that the maximum value of sin - 1 x , sin - 1 y , sin - 1 z is π 2 and minimum value of sin - 1 x , sin - 1 y , sin - 1 z is π 2 Now, For maximum value LHS = sin - 1 x 2 + sin - 1 y 2 + sin - 1 z 2 = π 2 2 + π 2 2 + π 2 2 = 3 4 π 2 = RHS and For minimum value LHS = sin - 1 x 2 + sin - 1 y 2 + sin - 1 z 2 = - π 2 2 + - π 2 2 + - π 2 2 = 3 4 π 2 = RHS Now, For maximum value sin - 1 x = π 2 , sin - 1 y = π 2 , sin - 1 z = π 2 ⇒ x = sin π 2 , y = sin π 2 , z = sin π 2 ⇒ x = 1 , y = 1 , z = 1 ∴ x 2 + y 2 + z 2 = 1 + 1 + 1 = 3 and for minimum value sin - 1 x = - π 2 , sin - 1 y = - π 2 , sin - 1 z = - π 2 ⇒ x = sin - π 2 , y = sin - π 2 , z = sin - π 2 ⇒ x = - 1 , y = - 1 , z = - 1 ∴ x 2 + y 2 + z 2 = 1 + 1 + 1 = 3

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MATHEMATICS CLASS XII VOLUME-1>Chapter 4 - Inverse Trigonometric Functions>EXERCISE

R D Sharma Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: EXERCISE

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: EXERCISE

Attempt the free practice questions on Chapter 4: Inverse Trigonometric Functions, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. MATHEMATICS CLASS XII VOLUME-1 solutions are prepared by Experienced Embibe Experts.