Solve the following equation: 2 tan - 1 x = cos - 1 1 - a 2 1 + a 2 - cos - 1 1 - b 2 1 + b 2 a > 0 , b > 0

We have, 2 tan - 1 x = cos - 1 1 - a 2 1 + a 2 - cos - 1 1 - b 2 1 + b 2 where, a > 0 , b > 0 ⇒ 2 tan - 1 x = 2 tan - 1 a - 2 tan - 1 b ∵ 2 tan - 1 x = cos - 1 1 - x 2 1 + x 2 , x ≥ 0 ⇒ tan - 1 x = tan - 1 a - tan - 1 b ⇒ tan - 1 x = tan - 1 a - b 1 + a b ⇒ x = a - b 1 + a b

Solve the equation cos - 1 x 2 - 1 x 2 + 1 + tan - 1 2 x x 2 - 1 = 2 π 3 .

Given that cos - 1 x 2 - 1 x 2 + 1 + tan - 1 2 x x 2 - 1 = 2 π 3 Let cos - 1 x 2 - 1 x 2 + 1 = θ or cos θ = x 2 - 1 x 2 + 1 ⇒ tan θ = 1 - cos 2 θ cos θ = x 2 + 1 - x 2 - 1 2 x 2 - 1 = 2 x x 2 - 1 Or θ = tan - 1 ⁡ 2 x x 2 - 1 Given, cos - 1 x 2 - 1 x 2 + 1 + tan - 1 2 x x 2 - 1 = 2 π 3 ⇒ tan - 1 ⁡ 2 x x 2 - 1 + tan - 1 ⁡ 2 x x 2 - 1 = 2 π 3 ∵ tan - 1 A + tan - 1 B = tan - 1 A + B 1 - A B , A B < 1 and A , B > 0 ⇒ tan - 1 2 x x 2 - 1 = π 3 ⇒ 2 x x 2 - 1 = tan π 3 = 3 ⇒ 3 x 2 - 2 x - 3 = 0 ⇒ x 2 3 - 3 x + x - 3 = 0 ⇒ x 3 + 1 x - 3 = 0 ∴ x = 3 , - 1 3

Find the sum of the series: tan - 1 1 3 + tan - 1 2 9 + . . . . + tan - 1 2 n - 1 1 + 2 2 n - 1 + . . . . ∞

We have, tan - 1 1 3 + tan - 1 2 9 + … + tan - 1 2 n - 1 1 + 2 2 n - 1 + . . . ∞ Then, T n = tan - 1 2 n - 1 1 + 2 2 n - 1 ⇒ T n = tan - 1 2 n - 2 n - 1 1 + 2 n 2 n - 1 ⇒ T n = tan - 1 2 n - tan - 1 2 n - 1 Now, T 1 = tan - 1 2 - tan - 1 1 T 2 = tan - 1 2 2 - tan - 1 2 T 3 = tan - 1 2 3 - tan - 1 2 2 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ T n = tan - 1 2 n - tan - 1 2 n - 1 S n = tan - 1 2 n - tan - 1 1 ⇒ S ∞ = lim n → ∞ tan - 1 2 n - π 4 ⇒ S ∞ = π 2 - π 4 = π 4

Find the sum of the series: tan - 1 1 x 2 + x + 1 + tan - 1 1 x 2 + 3 x + 3 + tan - 1 1 x 2 + 5 x + 7 + tan - 1 1 x 2 + 7 x + 13 ⋯ ⋯ ⋯ upto n terms

S n = tan - 1 x + 1 - x 1 + x x + 1 + tan - 1 x + 2 - x + 1 1 + x + 2 x + 1 + tan - 1 x + 3 - x + 2 1 + x + 3 x + 2 · · · · · · · · · · · · tan - 1 x + n - x + n - 1 1 + x + n x + n - 1 = tan - 1 x + 1 - tan - 1 x + tan - 1 x + 2 - tan - 1 x + 1 + tan - 1 x + 3 - tan - 1 x + 2 + · · · · · · · tan - 1 x + n - tan - 1 x + n - 1 Cancel out the similar term with negative sign, we get, S n = tan - 1 x + n - tan - 1 x .

Find the domain of definition the following functions. f x = sin cos x + ln - 2 cos 2 x + 3 cos x + 1 + e cos - 1 2 sin x + 1 2 2 sin x

We have, f x = sin cos x + ln - 2 cos 2 x + 3 cos x + 1 + e cos - 1 2 sin x + 1 2 2 sin x Here, f x is defined if i sin cos x ≥ 0 ⇒ cos ≥ 0 ⇒ x ∈ 2 n π - π 2 , 2 nπ + π 2 , n ∈ Z ii - 2 cos 2 x + 3 cos x + 1 > 0 ⇒ 2 cos 2 x - 3 cos x - 1 < 0 ⇒ cos x - 3 + 17 4 cos x - 3 - 17 4 < 0 ⇒ 3 - 17 4 < cos x < 3 + 17 4 By i ∩ ii , 0 ≤ cos x < 3 + 17 4 , hence x ∈ 2 n π - π 2 , 2 nπ + π 2 , n ∈ Z iii 2 sin x > 0 ⇒ sin x > 0 ⇒ x ∈ 2 n π , 2 n + 1 π , n ∈ Z And, - 1 ≤ 2 sin x + 1 2 2 sin x ≤ 1 ⇒ 2 sin x + 1 ≥ - 2 2 sin x ⇒ 2 sin x + 2 2 sin x + 1 ≥ 0 ⇒ 2 sin x + 1 2 ≥ 0 which is always true. And, 2 sin x + 1 2 2 sin x ≤ 1 ⇒ 2 sin x - 2 2 sin x + 1 ≤ 0 ⇒ 2 sin x - 1 2 ≤ 0 which is true, if 2 sin x - 1 = 0 ⇒ sin x = 1 2 ⇒ sin x = 1 2 = sin π 6 ⇒ x ∈ 2 n π + π 6 , 2 nπ + 5 π 6 , n ∈ Z Hence, i ∩ ii ∩ iii x ∈ 2 n π + π 6 ; n ∈ Z

Prove that sin - 1 cos sin - 1 x + cos - 1 sin cos - 1 x = π 2 , | x | ≤ 1 .

We have to prove that sin - 1 cos sin - 1 x + cos - 1 sin cos - 1 x = π 2 Let p = sin - 1 x Then, sin - 1 cos sin - 1 x + cos - 1 sin cos - 1 x = sin - 1 cos p + cos - 1 sin π 2 - p ∵ sin - 1 x + cos - 1 x = π 2 = sin - 1 cos p + cos - 1 cos p = π 2 Hence, proved

E M B I B E

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-4>Q 23

MEDIUM

JEE Main/Advance

IMPORTANT

Earn 100

Solve the following equation:

$\sin^{- 1} x + \sin^{- 1} (1 - x) = \cos^{- 1} x$

Important Questions on Inverse Trigonometric Functions

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-4>Q 24

Solve the following equation:

$2 \tan^{- 1} x = \cos^{- 1} \frac{1 - a^{2}}{1 + a^{2}} - \cos^{- 1} \frac{1 - b^{2}}{1 + b^{2}} a > 0, b > 0$

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-4>Q 25

Solve the equation

\cos^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) + \tan^{- 1} (\frac{2 x}{x^{2} - 1}) = \frac{2 π}{3} .

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-4>Q 26

Find the sum of the series:

$\tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{2}{9} + . . . . + \tan^{- 1} \frac{2^{n - 1}}{1 + 2^{2 n - 1}} + . . . . \infty$

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-4>Q 27

Find the sum of the series:

$\cot^{- 1} 7 + \cot^{- 1} 13 + \cot^{- 1} 21 + \cot^{- 1} 31 + . . .$ to $n$ terms.

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 15 - Inverse Trigonometric Functions>EXERCISE-4>Q 28

Find the sum of the series:

$\tan^{- 1} \frac{1}{x^{2} + x + 1} + \tan^{- 1} \frac{1}{x^{2} + 3 x + 3} + \tan^{- 1} \frac{1}{x^{2} + 5 x + 7} + \tan^{- 1} \frac{1}{x^{2} + 7 x + 13} \dots \dots \dots$ upto $n$ terms