Prove that tan - 1 1 2 + tan - 1 2 11 = tan - 1 3 4 .

To prove tan - 1 1 2 + tan - 1 2 11 = tan - 1 3 4 Taking L . H . S . = tan - 1 1 2 + tan - 1 2 11 = tan - 1 1 2 + 2 11 1 - 1 2 × 2 11 ∵ tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y = tan - 1 11 + 4 22 1 - 1 11 = tan - 1 15 22 10 11 = tan - 1 3 4 = R . H . S . Hence, proved.

If A = sin - 1 2 x 1 + x 2 , B = cos - 1 1 - x 2 1 + x 2 and C = tan - 1 2 x 1 - x 2 satisfies the condition 3 A - 4 B + 2 C = π 3 . Find the value of x .

Given A = sin - 1 2 x 1 + x 2 , B = cos - 1 1 - x 2 1 + x 2 and C = tan - 1 2 x 1 - x 2 Given that they satisfy the condition 3 A - 4 B + 2 C = π 3 ⇒ 3 sin - 1 2 x 1 + x 2 - 4 cos - 1 1 - x 2 1 + x 2 + 2 tan - 1 2 x 1 - x 2 = π 3 Put x = tan y ⇒ 3 sin - 1 2 tan y 1 + tan 2 y - 4 cos - 1 1 - tan 2 y 1 + tan 2 y + 2 tan - 1 2 tan y 1 - tan 2 y = π 3 ⇒ 3 sin - 1 sin 2 y - 4 cos - 1 cos 2 y + 2 tan - 1 tan 2 y = π 3 ⇒ 3 2 y - 4 2 y + 2 2 y = π 3 ⇒ 6 y - 8 y + 4 y = π 3 ⇒ 2 y = π 3 ⇒ y = π 6 ⇒ tan - 1 x = π 6 ⇒ x = tan π 6 = 1 3 .

Solve tan - 1 2 x + tan - 1 3 x = π 4 .

Given that, tan - 1 2 x + tan - 1 3 x = π 4 ⇒ tan - 1 2 x + 3 x 1 - 2 x · 3 x = π 4 ⇒ 5 x 1 - 6 x 2 = tan π 4 ⇒ 5 x 1 - 6 x 2 = 1 ⇒ 5 x = 1 - 6 x 2 ⇒ 6 x 2 + 5 x - 1 = 0 ⇒ 6 x 2 + 6 x - x - 1 = 0 ⇒ 6 x ( x + 1 ) - 1 ( x + 1 ) = 0 ⇒ ( x + 1 ) ( 6 x - 1 ) = 0 ⇒ x + 1 = 0 and 6 x - 1 = 0 ⇒ x = - 1 and x = 1 6 But x = 1 6 satisfies the given equation ∴ x = 1 6

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 2 - Inverse Trigonometric Functions>Kerala Board-2018

Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: Kerala Board-2018

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: Kerala Board-2018

Attempt the free practice questions on Chapter 2: Inverse Trigonometric Functions, Exercise 1: Kerala Board-2018 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: Kerala Board-2018 with Hints & Solutions

EASY

12th Kerala Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 2 - Inverse Trigonometric Functions>Kerala Board-2018>Q 1

Question Image

Identify the function from the above graph.

EASY

12th Kerala Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 2 - Inverse Trigonometric Functions>Kerala Board-2018>Q 2

Question Image

Find the domain and range of the function represented in above graph.

EASY

12th Kerala Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 2 - Inverse Trigonometric Functions>Kerala Board-2018>Q 3

Prove that $\tan^{- 1} \frac{1}{2} + \tan^{- 1} \frac{2}{11} = \tan^{- 1} \frac{3}{4}$ .

HARD

12th Kerala Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 2 - Inverse Trigonometric Functions>Kerala Board-2019>Q 4

If $A = \sin^{- 1} (\frac{2 x}{1 + x^{2}}), B = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})$ and $C = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ satisfies the condition $3 A - 4 B + 2 C = \frac{π}{3}$ . Find the value of $x$ .