Solve the following equation for x : tan - 1 1 - x 1 + x - 1 2 tan - 1 x = 0 , where x > 0

Given that, tan - 1 1 - x 1 + x - 1 2 tan - 1 ( x ) = 0 ⇒ tan - 1 1 - x 1 + x = 1 2 tan - 1 ( x ) ⇒ tan - 1 1 - tan - 1 x = 1 2 tan - 1 ( x ) ∵ tan - 1 1 - tan - 1 x = tan - 1 1 - x 1 + x ⇒ tan - 1 1 = 3 2 tan - 1 ( x ) ⇒ π 4 = 3 2 tan - 1 ( x ) ⇒ π 6 = tan - 1 ( x ) ⇒ x = 1 3

Solve the following equation for x : cot - 1 x - cot - 1 ( x + 2 ) = π 12 , where x > 0

Given that, cot - 1 ( x ) - cot - 1 ( x + 2 ) = π 12 ⇒ tan - 1 1 x - tan - 1 1 x + 2 = π 12 ∵ cot - 1 x = tan - 1 1 x ⇒ tan - 1 1 x - 1 x + 2 1 + 1 x ( x + 2 ) = π 12 ⇒ tan - 1 2 x ( x + 2 ) x 2 + 2 x + 1 x ( x + 2 ) = π 12 ⇒ tan - 1 2 x 2 + 2 x + 1 = π 12 ⇒ 2 x 2 + 2 x + 1 = tan π 12 ⇒ 2 x 2 + 2 x + 1 = tan π 3 - π 4 ⇒ 2 x 2 + 2 x + 1 = tan π 3 - tan π 4 1 + tan π 3 × tan π 4 ⇒ 2 x 2 + 2 x + 1 = 3 - 1 3 + 1 ⇒ 2 x 2 + 2 x + 1 = 3 - 1 3 + 1 × 3 + 1 3 + 1 ⇒ 2 x 2 + 2 x + 1 = 2 ( 3 + 1 ) 2 ⇒ 1 ( x + 1 ) 2 = 1 ( 3 + 1 ) 2 ⇒ x + 1 = 3 + 1 ⇒ x = 3

Solve the following equation for x : tan - 1 x 2 + tan - 1 x 3 = π 4 , 0 < x < 6

Given that, tan - 1 x 2 + tan - 1 x 3 = π 4 , 0 < x < 6 We know that, tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y , x y < 1 ⇒ tan - 1 x 2 + x 3 1 - x 2 × x 3 = π 4 ⇒ tan - 1 5 x 6 6 - x 2 6 = π 4 ⇒ 5 x 6 - x 2 = tan π 4 ⇒ 5 x 6 - x 2 = 1 ⇒ x 2 + 5 x - 6 = 0 ⇒ ( x - 1 ) ( x + 6 ) = 0 ⇒ x = 1 [ ∵ 0 < x < 6 ]

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

We know tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y tan - 1 x - 2 x - 4 + tan - 1 x + 2 x + 4 = π 4 ⇒ tan - 1 x - 2 x - 4 + x + 2 x + 4 1 - x - 2 x - 4 × x + 2 x + 4 = π 4 ⇒ tan - 1 x 2 + 2 x - 8 + x 2 - 2 x - 8 ( x - 4 ) ( x + 4 ) x 2 - 16 - x 2 + 4 ( x - 4 ) ( x + 4 ) = π 4 ⇒ 2 x 2 - 16 - 12 = tan π 4 ⇒ 2 x 2 - 16 - 12 = 1 ⇒ 2 x 2 - 16 = - 12 ⇒ 2 x 2 = 4 ⇒ x 2 = 2 ⇒ x = ± 2

Solve the following equation for x : tan - 1 ( 2 + x ) + tan - 1 ( 2 - x ) = tan - 1 2 3 , where x < - 3 or, x > 3 .

Given that, tan - 1 ( 2 + x ) + tan - 1 ( 2 - x ) = tan - 1 2 3 We know that, tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y , x y < 1 ⇒ tan - 1 2 + x + 2 - x 1 - ( 2 + x ) × ( 2 - x ) = tan - 1 2 3 ⇒ 4 1 - 4 + x 2 = 2 3 ⇒ 2 x 2 = 18 ⇒ x 2 = 9 ⇒ x = ± 3

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

Given that, tan - 1 x - 2 x - 1 + tan - 1 x + 2 x + 1 = π 4 ⇒ tan - 1 x - 2 x - 1 + x + 2 x + 1 1 - x - 2 x - 1 x + 2 x + 1 = π 4 tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y ⇒ ( x - 2 ) ( x + 1 ) + ( x - 1 ) ( x + 2 ) ( x - 1 ) ( x + 1 ) ( x - 1 ) ( x + 1 ) - ( x - 2 ) ( x + 2 ) ( x - 1 ) ( x + 1 ) = tan π 4 ⇒ ( x - 2 ) ( x + 1 ) + ( x - 1 ) ( x + 2 ) ( x - 1 ) ( x + 1 ) - ( x - 2 ) ( x + 2 ) = 1 ⇒ x 2 - x - 2 + x 2 + x - 2 x 2 - 1 - x 2 - 4 = 1 ⇒ 2 x 2 - 4 3 = 1 ⇒ 2 x 2 - 4 = 3 ⇒ 2 x 2 = 7 ⇒ x 2 = 7 2 x = ± 7 2

Sum the following series: tan - 1 1 3 + tan - 1 2 9 + tan - 1 4 33 + . . . + tan - 1 2 n - 1 1 + 2 2 n - 1

Given: tan - 1 1 3 + tan - 1 2 9 + tan - 1 4 33 + . . . + tan - 1 2 n - 1 1 + 2 2 n - 1 ⇒ tan - 1 2 - 1 1 + 2 × 1 + tan - 1 4 - 2 1 + 4 × 2 + tan - 1 8 - 4 1 + 8 × 4 + . . . + tan - 1 2 n - 2 n - 1 1 + 2 n × 2 n - 1 ⇒ tan - 1 2 - tan - 1 1 + tan - 1 4 - tan - 1 2 + tan - 1 8 - tan - 1 4 + . . . . . . . + tan - 1 2 n - tan - 1 2 n - 1 ⇒ - tan - 1 1 + tan - 1 2 n ⇒ tan - 1 2 n - tan - 1 1 Thus, sum of the series tan - 1 1 3 + tan - 1 2 9 + tan - 1 4 33 + . . . + tan - 1 2 n - 1 1 + 2 2 n - 1 = tan - 1 2 n - π 4 .

E M B I B E

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.

Questions from R D Sharma Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: EXERCISE with Hints & Solutions

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 4 - Inverse Trigonometric Functions>EXERCISE>Q 3

Solve the following equation for $x$ : $\Rightarrow \frac{π}{4} = \frac{3}{2} \tan^{- 1} (x)$ where $x > 0$

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 4 - Inverse Trigonometric Functions>EXERCISE>Q 3

Solve the following equation for $x :$ $\cot^{- 1} x - \cot^{- 1} (x + 2) = \frac{π}{12},$ where $x > 0$

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 4 - Inverse Trigonometric Functions>EXERCISE>Q 3

Solve the following equation for $x$ : $\tan^{- 1} (x + 2) + \tan^{- 1} (x - 2) = \tan^{- 1} (\frac{8}{79}), x > 0$

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 4 - Inverse Trigonometric Functions>EXERCISE>Q 3

Solve the following equation for $x$ : $\tan^{- 1} \frac{x}{2} + \tan^{- 1} \frac{x}{3} = \frac{π}{4}, 0 < x < \sqrt{6}$

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 4 - Inverse Trigonometric Functions>EXERCISE>Q 3

Solve the following equation forx:
$\tan^{- 1} (\frac{x - 2}{x - 4}) + \tan^{- 1} (\frac{x + 2}{x + 4}) = \frac{π}{4}$

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 4 - Inverse Trigonometric Functions>EXERCISE>Q 3

Solve the following equation for x : $\tan^{- 1} (2 + x) + \tan^{- 1} (2 - x) = \tan^{- 1} \frac{2}{3},$ where $x < - \sqrt{3}$ or, $x > \sqrt{3}$ .

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 4 - Inverse Trigonometric Functions>EXERCISE>Q 3

Solve the following equation for $x$ : $\tan^{- 1} \frac{x - 2}{x - 1} + \tan^{- 1} \frac{x + 2}{x + 1} = \frac{π}{4}$

HARD

12th CBSE

IMPORTANT

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

Sum the following series:
$\tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{2}{9} + \tan^{- 1} \frac{4}{33} + . . . + \tan^{- 1} \frac{2^{n - 1}}{1 + 2^{2 n - 1}}$

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