Find the real solutions of the equation tan - 1 x ( x + 1 ) + sin - 1 x 2 + x + 1 = π 2 .

We have to find the real solutions of the equation, tan - 1 x ( x + 1 ) + sin - 1 x 2 + x + 1 = π 2 . . … i Let sin - 1 x 2 + x + 1 = θ ⇒ sin θ = x 2 + x + 1 1 ⇒ tan θ = x 2 + x + 1 - x 2 - x ∵ tan θ = sin θ cos θ ∵ θ = tan - 1 x 2 + x + 1 - x 2 - x = sin - 1 x 2 + x + 1 On putting the value of θ in Eq. ( i ) , we get tan - 1 x ( x + 1 ) + tan - 1 x 2 + x + 1 - x 2 - x = π 2 We know that, tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y ; x y < 1 tan - 1 x ( x + 1 ) + x 2 + x + 1 - x 2 - x 1 - x ( x + 1 ) x 2 + x + 1 - x 2 - x = π 2 tan - 1 x 2 + x + x 2 + x + 1 - 1 x 2 + x 1 - x 2 + x · x 2 + x + 1 - 1 x 2 + x = x 2 ⇒ x 2 + x + - x 2 + x + 1 [ 1 - - x 2 + x + 1 x 2 + x ] = tan π 2 = 1 0 ⇒ 1 - - x 2 + x + 1 x 2 + x = 0 ⇒ - x 2 + x + 1 = 1 or x 2 + x = 0 ⇒ - x 2 - x - 1 = 1 or x ( x + 1 ) = 0 ⇒ x 2 + x + 2 = 0 or x ( x + 1 ) = 0 ∴ x = - 1 ± 1 - 4 × 2 2 = - 1 ± 7 i 2 ⇒ x = 0 or x = - 1 Hence, the real solutions of the equation are x = 0 , - 1 .

Solve the following equation cos tan - 1 x = sin cot - 1 3 4 .

We have to solve the equation: cos tan - 1 x = sin cot - 1 3 4 Let tan - 1 x = θ 1 ⇒ tan θ 1 = x 1 ⇒ cos θ 1 = 1 x 2 + 1 ⇒ θ 1 = cos - 1 1 x 2 + 1 And cot - 1 3 4 = θ 2 ⇒ cot θ 2 = 3 4 ⇒ sin θ 2 = 4 5 ⇒ θ 2 = sin - 1 4 5 ⇒ cos cos - 1 1 x 2 + 1 = sin sin - 1 4 5 ⇒ 1 x 2 + 1 = 4 5 ⇒ 4 x 2 + 1 = 5 ∵ cos cos - 1 x = x , x ∈ [ - 1 , 1 ] and ∵ sin sin - 1 x = x , x ∈ [ - 1 , 1 ] On squaring both the sides, we get ⇒ 16 x 2 + 1 = 25 ⇒ 16 x 2 = 9 ⇒ x 2 = 3 4 2 ∴ x = ± 3 4 = - 3 4 , 3 4 .

Find the simplified form of cos - 1 3 5 cos x + 4 5 sin x , where x ∈ - 3 π 4 , π 4 .

We have to find the simplified form of cos - 1 3 5 cos x + 4 5 sin x , x ∈ - 3 π 4 , π 4 Let cos y = 3 5 ⇒ sin y = 4 5 ⇒ y = cos - 1 3 5 = sin - 1 4 5 = tan - 1 4 3 ∴ cos - 1 3 5 cos x + 4 5 sin x = cos - 1 [ cos y · cos x + sin y · sin x ] cos - 1 [ cos y · cos x + sin y · sin x ] = cos - 1 [ cos ( y - x ) ] [ ∴ cos ( A - B ) = cos A cos B + sin A · sin B ] cos - 1 [ cos ( y - x ) ] = y - x ∵ cos - 1 cos x = x y - x = tan - 1 4 3 - x ∵ y = tan - 1 4 3 Hence, the simplified form is tan - 1 4 3 - x .

E M B I B E

NCERT Exemplar Mathematics - Class 12>Chapter 2 - Inverse Trigonometric Functions>EXERCISE

NCERT Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: EXERCISE

Author:NCERT

NCERT Mathematics Solutions for Exercise - NCERT Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: EXERCISE

Attempt the practice questions on Chapter 2: Inverse Trigonometric Functions, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. NCERT Exemplar Mathematics - Class 12 solutions are prepared by Experienced Embibe Experts.

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

HARD

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 2 - Inverse Trigonometric Functions>EXERCISE>Q 1

Find the value of $\tan^{- 1} (\tan \frac{5 π}{6}) + \cos^{- 1} (\cos \frac{13 π}{6})$ .

EASY

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 2 - Inverse Trigonometric Functions>EXERCISE>Q 2

Evaluate: $\cos [\cos^{- 1} (\frac{- \sqrt{3}}{2}) + \frac{π}{6}]$

HARD

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 2 - Inverse Trigonometric Functions>EXERCISE>Q 7

Find the real solutions of the equation $\tan^{- 1} \sqrt{x (x + 1)} + \sin^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}$ .

MEDIUM

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 2 - Inverse Trigonometric Functions>EXERCISE>Q 11

Solve the following equation $\cos (\tan^{- 1} x) = \sin (\cot^{- 1} \frac{3}{4})$ .

MEDIUM

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 2 - Inverse Trigonometric Functions>EXERCISE>Q 13

Find the simplified form of $\cos^{- 1} (\frac{3}{5} \cos x + \frac{4}{5} \sin x)$ , where $x \in [\frac{- 3 π}{4}, \frac{π}{4}]$ .

HARD

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 2 - Inverse Trigonometric Functions>EXERCISE>Q 17

If the value of $4 \tan^{- 1} \frac{1}{5} - \tan^{- 1} \frac{1}{239}$ is $\frac{π}{k}$ , then find the value of $k$ .

MEDIUM

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 2 - Inverse Trigonometric Functions>EXERCISE>Q 19

If $a_{1}, a_{2}, a_{3}, . . . . a_{n}$ is an arithmetic progression with common difference $d,$ then evaluate the following expression.

$\tan [\tan^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + \tan^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + \tan^{- 1} (\frac{d}{1 + a_{3} a_{4}}) + \dots + \tan^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})]$