Solve the following equation: sin - 1 x + sin - 1 2 x = π 3

We have, sin - 1 x + sin - 1 2 x = π 3 ⇒ s i n - 1 x 1 - ( 2 x ) 2 + 2 x 1 - x 2 = π 3 ∵ sin - 1 x + sin - 1 y = sin - 1 x 1 - y 2 + y 1 - x 2 ⇒ s i n - 1 x 1 - 4 x 2 + 2 x 1 - x 2 = π 3 ⇒ x 1 - 4 x 2 + 2 x 1 - x 2 = s i n π 3 ⇒ x 1 - 4 x 2 + 2 x 1 - x 2 = 3 2 ⇒ 2 x 1 - 4 x 2 + 4 x 1 - x 2 = 3 ⇒ 2 x 1 - 4 x 2 = 3 - 4 x 1 - x 2 Squaring both sides, ⇒ 4 x 2 1 - 4 x 2 = 3 + 16 x 2 1 - x 2 - 8 3 x 1 - x 2 ⇒ 8 3 x 1 - x 2 = 16 x 2 1 - x 2 + 3 - 4 x 2 ( 1 - 4 x 2 ) ⇒ 8 3 x 1 - x 2 = 16 x 2 - 16 x 4 + 3 - 4 x 2 + 16 x 4 ⇒ 8 3 x 1 - x 2 = 12 x 2 + 3 Again squaring both sides, ⇒ 192 x 2 ( 1 - x 2 ) = 12 x 2 + 3 2 ⇒ 192 x 2 - 192 x 4 = 144 x 4 + 9 + 72 x 2 ⇒ 336 x 4 - 120 x 2 + 9 = 0 ⇒ 112 x 4 - 40 x 2 + 3 = 0 ⇒ 112 x 4 - 28 x 2 - 12 x 2 + 3 = 0 ⇒ 28 x 2 4 x 2 - 1 - 3 4 x 2 - 1 = 0 ⇒ 28 x 2 - 3 4 x 2 - 1 = 0 ⇒ x 2 = 3 28 or x 2 = 1 4 ⇒ x = ± 3 28 , ± 1 2

If cos - 1 x + cos - 1 y = π 2 , then prove that cos - 1 x = sin - 1 y .

From properties of inverse trigonometric functions for complementary angles, sin - 1 y + cos - 1 y = π 2 , x ∈ - 1 , 1 ⇒ sin - 1 y = π 2 - cos - 1 y Given cos - 1 x + cos - 1 y = π 2 ⇒ cos - 1 x = π 2 - cos - 1 y ⇒ cos - 1 x = sin - 1 y Hence, proved.

Prove that cos - 1 - x = π - cos - 1 x , x ∈ - 1 , 1 .

Let cos A = x , x ∈ - 1 , 1 then, cos π - A = - cos A ⇒ cos π - A = - x So, cos - 1 x = A . . . . i And cos - 1 - x = π - A . . . . . ii Adding equations i and ii , cos - 1 x + cos - 1 - x = A + π - A ⇒ cos - 1 x + cos - 1 - x = π ⇒ cos - 1 - x = π - cos - 1 x

Prove that 2 tan - 1 1 2 + tan - 1 1 7 = tan - 1 31 17 .

Consider the given equation: 2 tan - 1 1 2 + tan - 1 1 7 = tan - 1 31 17 L . H . S . = 2 tan - 1 1 2 + tan - 1 1 7 Using 2 tan - 1 x = tan - 1 2 x 1 - x 2 = tan - 1 2 × 1 2 1 - 1 2 2 + tan - 1 1 7 = tan - 1 1 4 - 1 4 + tan - 1 1 7 = tan - 1 4 3 + tan - 1 1 7 Using tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y ; x , y > 0 and x y < 1 = tan - 1 4 3 + 1 7 1 - 4 3 × 1 7 = tan - 1 28 + 3 21 21 - 4 21 = tan - 1 31 21 17 21 = tan - 1 31 17 = R . H . S . Hence, proved.

Find the principal value of sin - 1 sin 7 π 4 .

We know that sin - 1 sin x = x , x ∈ - π 2 , π 2 Given sin - 1 sin 7 π 4 = sin - 1 sin 2 π - π 4 ⇒ sin - 1 sin 7 π 4 = sin - 1 - sin π 4 ⇒ sin - 1 sin 7 π 4 = sin - 1 sin - π 4 ⇒ sin - 1 sin 7 π 4 = - π 4

If cos - 1 x a + cos - 1 y b = θ , then prove that x 2 a 2 - 2 x y a b cos θ + y 2 b 2 = sin 2 θ .

Given that, cos - 1 x a + cos - 1 y b = θ and To prove: x 2 a 2 - 2 x y a b cos θ + y 2 b 2 = sin 2 θ Using cos - 1 x + cos - 1 y = cos - 1 x y - 1 - x 2 1 - y 2 . cos - 1 x a + cos - 1 y b = θ ⇒ cos - 1 x a × y b - 1 - x a 2 1 - y b 2 = θ ⇒ x a × y b - 1 - x a 2 1 - y b 2 = cos θ ⇒ x y a b - 1 - x 2 a 2 1 - y 2 b 2 = cos θ ⇒ x y a b - cos θ = 1 - x 2 a 2 1 - y 2 b 2 ⇒ x y a b - cos θ 2 = 1 - x 2 a 2 1 - y 2 b 2 2 ⇒ x y a b 2 + cos 2 θ - 2 x y a b cos θ = 1 - x 2 a 2 1 - y 2 b 2 ⇒ x y a b 2 + cos 2 θ - 2 x y a b cos θ = 1 - y 2 b 2 - x 2 a 2 + x y a b 2 ⇒ y 2 b 2 + x 2 a 2 - 2 x y a b cos θ = 1 - cos 2 θ ⇒ y 2 b 2 + x 2 a 2 - 2 x y a b cos θ = sin 2 θ ∵ cos 2 x + sin 2 x = 1 , x ∈ R Hence, proved.

Solve the equation sec - 1 x a - sec - 1 x b = sec - 1 b - sec - 1 a .

Given that, sec - 1 x a - sec - 1 x b = sec - 1 b - sec - 1 a We know that, sec - 1 x = cos - 1 1 x So, cos - 1 a x - cos - 1 b x = cos - 1 1 b - cos - 1 1 a Let cos - 1 a x = θ or cos θ = a x and sin θ = 1 x x 2 - a 2 cos - 1 b x = ϕ or cos ϕ = b x and sin ϕ = 1 x x 2 - b 2 cos - 1 1 b = α or cos α = 1 b and sin α = 1 b b 2 - 1 cos - 1 1 a = β or cos β = 1 a and sin β = 1 a a 2 - 1 Now, θ - ϕ = α - β or θ + β = α + ϕ ⇒ cos θ + β = cos α + ϕ ⇒ cos θ cos β - sin θ sin β = cos α cos ϕ - sin α sin ϕ ⇒ a x 1 a - 1 x x 2 - a 2 1 a a 2 - 1 = 1 b b x - 1 x x 2 - b 2 1 b b 2 - 1 ⇒ 1 x - 1 a x x 2 - a 2 a 2 - 1 = 1 x - 1 b x x 2 - b 2 b 2 - 1 ⇒ 1 a x x 2 - a 2 a 2 - 1 = 1 b x x 2 - b 2 b 2 - 1 ⇒ b a 2 x 2 - a 4 - x 2 + a 2 = a b 2 x 2 - b 4 - x 2 + b 2 ⇒ b 2 a 2 x 2 - a 4 - x 2 + a 2 = a 2 b 2 x 2 - b 4 - x 2 + b 2 ⇒ b 2 a 2 x 2 - b 2 a 4 - b 2 x 2 + b 2 a 2 = a 2 b 2 x 2 - a 2 b 4 - a 2 x 2 + a 2 b 2 ⇒ a 2 x 2 - b 2 x 2 = b 2 a 4 - a 2 b 4 ⇒ x 2 a 2 - b 2 = a 2 b 2 a 2 - b 2 ⇒ x 2 = a 2 b 2 ⇒ x = ± a b

E M B I B E

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

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

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

Attempt the free practice questions on Chapter 2: Inverse Trigonometric Functions, Exercise 1: Uttar Pradesh 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: Uttar Pradesh Board-2018 with Hints & Solutions

MEDIUM

12th Uttar Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 2 - Inverse Trigonometric Functions>Uttar Pradesh Board-2018>Q 4

If $\cos^{- 1} + \cos^{- 1} y + \cos^{- 1} z = π$ , then prove that $x^{2} + y^{2} + z^{2} + 2 x y z = 1$ .

HARD

12th Uttar Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 2 - Inverse Trigonometric Functions>Uttar Pradesh Board-2018>Q 5

Solve the following equation:

$\sin^{- 1} (x) + \sin^{- 1} (2 x) = \frac{π}{3}$

EASY

12th Uttar Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 2 - Inverse Trigonometric Functions>Uttar Pradesh Board-2018>Q 6

If $\cos^{- 1} (x) + \cos^{- 1} (y) = \frac{π}{2}$ , then prove that $\cos^{- 1} (x) = \sin^{- 1} (y)$ .

EASY

12th Uttar Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 2 - Inverse Trigonometric Functions>Uttar Pradesh Board-2019>Q 7

Prove that $\cos^{- 1} (- x) = π - \cos^{- 1} (x), x \in [- 1, 1]$ .

MEDIUM

12th Uttar Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 2 - Inverse Trigonometric Functions>Uttar Pradesh Board-2019>Q 8

Prove that $2 \tan^{- 1} \frac{1}{2} + \tan^{- 1} \frac{1}{7} = \tan^{- 1} \frac{31}{17}$ .

EASY

12th Uttar Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 2 - Inverse Trigonometric Functions>Uttar Pradesh Board-2020>Q 9

Find the principal value of $\sin^{- 1} (\sin \frac{7 π}{4})$ .

HARD

12th Uttar Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 2 - Inverse Trigonometric Functions>Uttar Pradesh Board-2020>Q 10

If $\cos^{- 1} \frac{x}{a} + \cos^{- 1} \frac{y}{b} = θ$ , then prove that $\frac{x^{2}}{a^{2}} - \frac{2 x y}{a b} \cos θ + \frac{y^{2}}{b^{2}} = \sin^{2} θ$ .