Write the following in the simplest form: cot - 1 a x 2 - a 2 , | x | > a

Given: cot - 1 a x 2 - a 2 , | x | > a Let x = a sec θ and θ = sec - 1 x a Now, cot - 1 a x 2 - a 2 = cot - 1 a a 2 sec 2 θ - a 2 = cot - 1 a a tan 2 θ = cot - 1 ( cot θ ) = θ Thus, the simplest form is sec - 1 x a .

Write the following in the simplest form: tan - 1 { x + 1 + x 2 } , x ∈ R

Let x = cot θ Now, tan - 1 { x + 1 + x 2 } = tan - 1 { cot θ + 1 + cot 2 θ } = tan - 1 { cot θ + cosec θ } = tan - 1 cos θ + 1 sin θ = tan - 1 2 cos 2 θ 2 2 sin θ 2 cos θ 2 = tan - 1 cot θ 2 = π 2 - θ 2 = tan - 1 tan π 2 - θ 2 Thus, the simplest form is π 2 - cot - 1 x 2 .

Write the following in the simplest form: tan - 1 { 1 + x 2 - x } , x ∈ R

Let x = cot θ Now, tan - 1 { 1 + x 2 - x } = tan - 1 { 1 + cot 2 θ - cot θ } = tan - 1 { cosec θ - cot θ } = tan - 1 { cosec θ - cot θ } = tan - 1 1 - cos θ sin θ = tan - 1 2 sin 2 θ 2 2 sin θ 2 cos θ 2 = tan - 1 tan θ 2 = θ 2 Thus, the simplest form is cot - 1 x 2 .

Write the following in the simplest form: tan - 1 1 + x 2 - 1 x , x ≠ 0

Let x = tan θ Now, tan - 1 1 + x 2 - 1 x = tan - 1 1 + tan 2 θ - 1 tan θ = tan - 1 sec 2 θ - 1 tan θ = tan - 1 sec θ - 1 tan θ = tan - 1 1 - cos θ sin θ = tan - 1 2 sin 2 θ 2 2 sin θ 2 cos θ 2 = tan - 1 tan θ 2 = θ 2 Thus, the simplest form is tan - 1 x 2 .

Write the following in the simplest form: tan - 1 a - x a + x , - a < x < a

Let x = a cos θ Now, tan - 1 a - x a + x = tan - 1 a - a cos θ a + a cos θ = tan - 1 1 - cos θ 1 + cos θ = tan - 1 2 sin 2 θ 2 2 cos 2 θ 2 = tan - 1 tan θ 2 = θ 2 = 1 2 cos - 1 x a

Write the following in the simplest form: sin - 1 x + 1 - x 2 2 , - 1 2 < x < 1 2

Let x = sin θ Now, sin - 1 x + 1 - x 2 2 = sin - 1 sin θ + 1 - sin 2 θ 2 = sin - 1 sin θ + cos θ 2 = sin - 1 1 2 sin θ + 1 2 cos θ = sin - 1 1 2 sin θ + 1 2 cos θ = sin - 1 sin θ + π 4 = θ + π 4 = π 4 + sin - 1 x

Write the following in the simplest form: sin - 1 1 + x + 1 - x 2 , 0 < x < 1

Let x = cos θ Now, sin - 1 1 + x + 1 - x 2 = sin - 1 1 + cos θ + 1 - cos θ 2 = sin - 1 2 cos 2 θ 2 + 2 sin 2 θ 2 2 = sin - 1 cos θ 2 + sin θ 2 2 = sin - 1 1 2 sin θ 2 + 1 2 cos θ 2 = sin - 1 sin θ 2 + π 4 = θ 2 + π 4 = cos - 1 x 2 + π 4 ∴ sin - 1 1 + x + 1 - x 2 = cos - 1 x 2 + π 4

Write the following in the simplest form: sin 2 tan - 1 1 - x 1 + x

Let x = cos θ Now, sin 2 tan - 1 1 - x 1 + x = sin 2 tan - 1 1 - cos θ 1 + cos θ = sin 2 tan - 1 2 sin 2 θ 2 2 cos 2 θ 2 = sin 2 tan - 1 tan θ 2 = sin θ = sin cos - 1 x = sin sin - 1 ( 1 - x 2 ) = 1 - x 2

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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.