Simplify the following: tan - 1 1 - cos x 1 + cos x

tan - 1 1 - cos x 1 + cos x = tan - 1 2 sin 2 x 2 2 cos 2 x 2 = tan - 1 tan 2 x 2 = tan - 1 tan x 2 = x 2

Prove that: sin - 1 3 5 + sin - 1 8 17 = cos - 1 36 85

L . H . S . = sin - 1 3 5 + sin - 1 8 17 Let, sin - 1 3 5 = α sin α = 3 5 , cos α = 1 - sin 2 α = 1 - 9 25 cos α = 4 5 And sin - 1 8 17 = β , sin β = 8 17 cos β = 1 - sin 2 β = 1 - 64 289 cos β = 15 17 We know that cos α + β = cos α cos β - sin α sin β = 4 5 × 15 17 - 3 5 × 8 17 cos α + β = 36 85 α + β = cos - 1 36 85 Now substituting values of α and β , sin - 1 3 5 + sin - 1 8 17 = cos - 1 36 85 Hence proved.

Prove that: sin - 1 3 5 + cos - 1 12 13 = cos - 1 33 65

L . H . S . = sin - 1 3 5 + cos - 1 12 13 Let, sin - 1 3 5 = α sin α = 3 5 , cos α = 1 - sin 2 α = 1 - 9 25 cos α = 4 5 And cos - 1 12 13 = β , cos β = 12 13 sin β = 1 - cos 2 β = 1 - 144 169 sin β = 5 13 We know that cos α + β = cos α cos β - sin α sin β = 4 5 × 12 13 - 3 5 × 5 13 cos α + β = 33 65 α + β = cos - 1 33 65 Now substituting values of α and β , sin - 1 3 5 + cos - 1 12 13 = cos - 1 33 65 Hence proved.

Solve the equation tan - 1 x - 1 x - 2 + tan - 1 x + 1 x + 2 = π 4 for x .

Given, tan - 1 x - 1 x - 2 + tan - 1 x + 1 x + 2 = π 4 ...(i) Let, tan - 1 x - 1 x - 2 = α , tan - 1 x + 1 x + 2 = β ⇒ tan α = x - 1 x - 2 , tan β = x + 1 x + 2 ∴ α + β = π 4 [ from (i) ] ⇒ tan α + β = tan π 4 ⇒ tan α + tan β 1 - tan α tan β = 1 ⇒ x - 1 x - 2 + x + 1 x + 2 1 - x 2 - 1 x 2 - 4 = 1 ⇒ ( x - 1 ) ( x + 2 ) + ( x + 1 ) ( x - 2 ) x 2 - 4 - x 2 + 1 = 1 ⇒ x 2 + x - 2 + x 2 - x - 2 - 3 = 1 ⇒ 2 x 2 - 4 = - 3 ⇒ 2 x 2 = 1 ∴ x = ± 1 2

Solve the equation tan - 1 1 2 x + 1 + tan - 1 1 4 x + 1 = tan - 1 2 x 2 for x .

Given, tan - 1 1 2 x + 1 + tan - 1 1 4 x + 1 = tan - 1 2 x 2 ...(i) Let, tan - 1 1 2 x + 1 = α , tan - 1 1 4 x + 1 = β ⇒ tan α = 1 2 x + 1 , tan β = 1 4 x + 1 ∴ α + β = tan - 1 2 x 2 [ from (i) ] ⇒ tan α + β = tan tan - 1 2 x 2 ⇒ tan α + tan β 1 - tan α tan β = 2 x 2 ⇒ 1 2 x + 1 + 1 4 x + 1 1 - 1 ( 2 x + 1 ) ( 4 x + 1 ) = 2 x 2 ⇒ 4 x + 1 + 2 x + 1 8 x 2 + 6 x + 1 - 1 = 2 x 2 ⇒ 6 x 3 + 2 x 2 = 16 x 2 + 12 x ⇒ 6 x 3 - 14 x 2 - 12 x = 0 ⇒ 2 x 3 x 2 - 7 x - 6 = 0 ⇒ 2 x 3 x 2 - 9 x + 2 x - 6 = 0 ⇒ 2 x ( 3 x + 2 ) ( x - 3 ) = 0 ∴ x = 0 , - 2 3 , 3

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Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 2 - Inverse Trigonometric Functions>Exercise

Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: Exercise

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: Exercise

Attempt the free practice questions on Chapter 2: Inverse Trigonometric Functions, Exercise 1: Exercise with hints and solutions to strengthen your understanding. Mathematics Crash Course (Based on Revised Syllabus-2023) solutions are prepared by Experienced Embibe Experts.