Find the value of sin - 1 4 5 + sin - 1 12 13 + sin - 1 33 65 .

Given that, sin - 1 4 5 + sin - 1 12 13 + sin - 1 33 65 = sin - 1 4 5 1 - 144 169 + 12 13 1 - 16 25 + sin - 1 33 65 = sin - 1 4 5 × 5 13 + 12 13 × 3 5 + sin - 1 33 65 = sin - 1 56 65 + cos - 1 1 - 33 65 2 = sin - 1 56 65 + cos - 1 4225 - 1089 4225 = sin - 1 56 65 + cos - 1 3136 4225 = sin - 1 56 65 + cos - 1 56 65 = π 2 [Using sin - 1 x + cos - 1 x = π 2 ; x ∈ [ - 1 , 1 ] ] Thus, sin - 1 4 5 + sin - 1 12 13 + sin - 1 33 65 = π 2 .

If sin - 1 x a + sin - 1 y b = sin - 1 c 2 a b , then prove that b 2 x 2 + 2 x y a 2 b 2 - c 4 + a 2 y 2 = c 4 .

Given that, sin - 1 x a + sin - 1 y b = sin - 1 c 2 a b and To prove: b 2 x 2 + 2 x y a 2 b 2 - c 4 + a 2 y 2 = c 4 Consider, sin - 1 x a + sin - 1 y b = sin - 1 c 2 a b ⇒ sin - 1 x a = sin - 1 c 2 a b - sin - 1 y b ⇒ x a = sin sin - 1 c 2 a b - sin - 1 y b ⇒ x a = sin sin - 1 c 2 a b cos sin - 1 y b - cos sin - 1 c a b sin sin - 1 y b [Using sin A - B = sin A cos B - cos A sin B ] ⇒ x a = c 2 a b cos cos - 1 1 - y 2 b 2 - cos cos - 1 1 - c 4 a 2 b 2 · y b [Using sin - 1 sin x = x ; x ∈ - π 2 , π 2 and sin - 1 x = cos - 1 1 - x 2 ] ⇒ x a = c 2 a b · 1 - y 2 b 2 - 1 - c 4 a 2 b 2 · y b ⇒ x a + y b 1 - c 4 a 2 b 2 = c 2 a b 1 - y 2 b 2 ⇒ x a + y b 1 - c 4 a 2 b 2 2 = c 2 a b 1 - y 2 b 2 2 ⇒ x 2 a 2 + y 2 b 2 1 - c 4 a 2 b 2 + 2 x y a b 1 - c 4 a 2 b 2 = c 4 a 2 b 2 1 - y 2 b 2 ⇒ x 2 a 2 + y 2 b 2 - y 2 b 2 · c 4 a 2 b 2 + 2 x y a b 1 - c 4 a 2 b 2 = c 4 a 2 b 2 - c 4 a 2 b 2 · y 2 b 2 ⇒ x 2 a 2 + y 2 b 2 + 2 x y a b 1 - c 4 a 2 b 2 = c 4 a 2 b 2 ⇒ x 2 a 2 + y 2 b 2 + 2 x y a 2 b 2 - c 4 a 2 b 2 = c 4 a 2 b 2 ⇒ b 2 x 2 + a 2 y 2 + 2 x y a 2 b 2 - c 4 = c 4 [Multiplied both sides with a 2 b 2 ] Hence, proved.

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

Given that, sin - 1 x a + sin - 1 y b = α and To prove: x 2 a 2 + 2 x y a b cos α + y 2 b 2 = sin 2 α Consider, sin - 1 x a + sin - 1 y b = α ⇒ sin - 1 x a = α - sin - 1 y b ⇒ x a = sin α - sin - 1 y b ⇒ x a = sin α · cos sin - 1 y b - cos α · sin sin - 1 y b [Using sin A - B = sin A cos B - cos A sin B ] ⇒ x a = sin α · cos cos - 1 1 - y 2 b 2 - cos α · y b [Using sin - 1 sin x = x ; x ∈ - π 2 , π 2 and sin - 1 x = cos - 1 1 - x 2 ] ⇒ x a = sin α · 1 - y 2 b 2 - cos α · y b ⇒ x a + cos α · y b = sin α · 1 - y 2 b 2 ⇒ x a + cos α · y b 2 = sin α · 1 - y 2 b 2 2 ⇒ x 2 a 2 + cos 2 α · y 2 b 2 + 2 x y a b cos α = sin 2 α · 1 - y 2 b 2 ⇒ x 2 a 2 + cos 2 α · y 2 b 2 + 2 x y a b cos α = sin 2 α - sin 2 α y 2 b 2 ⇒ x 2 a 2 + cos 2 α · y 2 b 2 + sin 2 α y 2 b 2 + 2 x y a b cos α = sin 2 α ⇒ x 2 a 2 + 2 x y a b cos α + y 2 b 2 = sin 2 α Hence, proved.

Solve the following equation: tan - 1 x - 1 x + 1 + tan - 1 2 x - 1 2 x + 1 = tan - 1 23 36 .

Given that, tan - 1 x - 1 x + 1 + tan - 1 2 x - 1 2 x + 1 = tan - 1 23 36 ⇒ tan - 1 x - 1 x + 1 + 2 x - 1 2 x + 1 1 - x - 1 x + 1 × 2 x - 1 2 x + 1 = tan - 1 23 36 [Using tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y ; x y < 1 ] ⇒ x - 1 x + 1 + 2 x - 1 2 x + 1 1 - x - 1 x + 1 × 2 x - 1 2 x + 1 = 23 36 ⇒ x - 1 2 x + 1 + 2 x - 1 x + 1 x + 1 2 x + 1 x + 1 2 x + 1 - x - 1 2 x - 1 x + 1 2 x + 1 = 23 36 ⇒ x - 1 2 x + 1 + 2 x - 1 x + 1 x + 1 2 x + 1 - x - 1 2 x - 1 = 23 36 ⇒ 2 x 2 + x - 2 x - 1 + 2 x 2 + 2 x - x - 1 2 x 2 + x + 2 x + 1 - 2 x 2 - x - 2 x + 1 = 23 36 ⇒ 4 x 2 - 2 6 x = 23 36 ⇒ 4 x 2 - 2 x = 23 6 ⇒ 24 x 2 - 12 = 23 x ⇒ 24 x 2 - 23 x - 12 = 0 ⇒ 24 x 2 - 32 x + 9 x - 12 = 0 ⇒ 8 x ( 3 x - 4 ) + 3 ( 3 x - 4 ) = 0 ⇒ ( 8 x + 3 ) ( 3 x - 4 ) = 0 ⇒ 8 x + 3 = 0 or 3 x - 4 = 0 ⇒ x = - 3 8 or x = 4 3 So, clearly x = - 3 8 does not satisfy the given equation. Thus, the required solution x = 4 3 .

Solve the following equation: tan - 1 1 3 + tan - 1 1 5 + tan - 1 1 7 + tan - 1 x = π 4 .

Given that, tan - 1 1 3 + tan - 1 1 5 + tan - 1 1 7 + tan - 1 x = π 4 ⇒ tan - 1 1 3 + tan - 1 1 5 + tan - 1 x = π 4 - tan - 1 1 7 ⇒ tan - 1 1 3 + tan - 1 1 5 + tan - 1 x = tan - 1 1 - tan - 1 1 7 ⇒ tan - 1 1 3 + 1 5 1 - 1 3 × 1 5 + tan - 1 x = tan - 1 1 - tan - 1 1 7 [Using tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y ; x y < 1 ] ⇒ tan - 1 5 + 3 15 15 - 1 15 + tan - 1 x = tan - 1 1 - 1 7 1 + 1 × 1 7 ⇒ tan - 1 8 14 + tan - 1 x = tan - 1 6 8 ⇒ tan - 1 4 7 + tan - 1 x = tan - 1 3 4 ⇒ tan - 1 x = tan - 1 3 4 - tan - 1 4 7 ⇒ tan - 1 x = tan - 1 3 4 - 4 7 1 + 3 4 × 4 7 ⇒ tan - 1 x = tan - 1 21 - 16 28 28 + 12 28 ⇒ tan - 1 x = tan - 1 5 40 ⇒ tan - 1 x = tan - 1 1 8 ⇒ x = 1 8 Thus, the required solution x = 1 8 .

E M B I B E

Elements of Mathematics Class 12>Chapter 2 - Inverse Trigonometric Functions>Exercises- 2

Odisha Board Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: Exercises- 2

Author:Odisha Board

Odisha Board Mathematics Solutions for Exercise - Odisha Board Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: Exercises- 2

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

Questions from Odisha Board Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: Exercises- 2 with Hints & Solutions

MEDIUM

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 2 - Inverse Trigonometric Functions>Exercises- 2>Q 10

Find the value of $\sin^{- 1} \frac{4}{5} + \sin^{- 1} \frac{12}{13} + \sin^{- 1} \frac{33}{65}$ .

HARD

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 2 - Inverse Trigonometric Functions>Exercises- 2>Q 12

If $\sin^{- 1} (\frac{x}{a}) + \sin^{- 1} (\frac{y}{b}) = \sin^{- 1} (\frac{c^{2}}{a b})$ , then prove that $b^{2} x^{2} + 2 x y \sqrt{a^{2} b^{2} - c^{4}} + a^{2} y^{2} = c^{4}$ .

HARD

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 2 - Inverse Trigonometric Functions>Exercises- 2>Q 12

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

HARD

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 2 - Inverse Trigonometric Functions>Exercises- 2>Q 12

If $\sin^{- 1} x + \sin^{- 1} y + \sin^{- 1} z = π$ , then prove that $x^{4} + y^{4} + z^{4} + 4 x^{2} y^{2} z^{2} = 2 (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2})$ .