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

We have to prove that: 2 tan - 1 1 2 + tan - 1 1 7 = tan - 1 31 17 Consider, LHS = 2 tan - 1 1 2 + tan - 1 1 7 Using: 2 tan - 1 x = tan - 1 2 x 1 - x 2 ; x ∈ ( - 1 , 1 ) = 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 = RHS Hence, proved.

Write cot - 1 1 x 2 - 1 , x > 1 in the simplest form.

The given function is, cot - 1 1 x 2 - 1 Let x = sec t ⇒ t = sec - 1 x = cot - 1 1 sec 2 t - 1 = cot - 1 1 tan 2 t ∵ tan 2 x = sec 2 x - 1 = cot - 1 1 tan t = cot - 1 cot t = t ∵ cot - 1 cot x = x = sec - 1 x [ Substituting value from t ] Hence, cot - 1 1 x 2 - 1 in simplest form is sec - 1 x .

Show that cos - 1 4 5 + cos - 1 12 13 = cos - 1 33 65 .

Consider, cos - 1 4 5 + cos - 1 12 13 = cos - 1 4 5 × 12 13 - 1 - 4 5 2 × 1 - 12 13 2 ∵ cos - 1 x + cos - 1 y = cos - 1 x y - 1 - x 2 · 1 - y 2 = cos - 1 48 65 - 1 - 16 25 × 1 - 144 169 = cos - 1 48 65 - 9 25 × 25 169 = cos - 1 48 65 - 3 5 × 5 13 = cos - 1 48 65 - 3 13 = cos - 1 48 - 15 65 = cos - 1 33 65 Hence, proved that cos - 1 4 5 + cos - 1 12 13 = cos - 1 33 65 .

Write the range of the function y = sec - 1 x

Given y = sec - 1 x ⇒ x = sec y Here, x ≥ 1 or x ≤ - 1 The domain includes all numbers from - ∞ to ∞ except the numbers between - 1 and 1 Therefore, range of sec - 1 x is all the angles between 0 and 180 ° except for 90 ° Therefore range = [ 0 , π ] - π 2

Prove that cot - 1 - x = π - cot - 1 x , ∀ x ∈ R .

To prove cot - 1 - x = π - cot - 1 x , ∀ x ∈ R Let y = cot - 1 - x Now apply the function ‘ cot ’ on both sides of the equation. ⇒ cot y = cot cot - 1 - x ⇒ cot y = - x ∵ cot cot - 1 x = x , x ∈ R ⇒ cot π - y = x ∵ cot π - x = - cot x Now again apply the inverse ‘ cot ’ function on both sides of the equation. ⇒ cot - 1 cot π - y = cot - 1 x ⇒ π - y = cot - 1 x ⇒ y = π - cot - 1 x Hence, proved.

Find the value of sin - 1 sin 3 π 5 .

We have to find the principal value of sin - 1 sin 3 π 5 Let y = sin - 1 sin 3 π 5 ⇒ y = sin - 1 sin π - 2 π 5 ⇒ y = sin - 1 sin 2 π 5 ∵ sin ( π - x ) = sin x We know that, sin - 1 ( sin x ) = x , x ∈ - π 2 , π 2 ∴ y = sin - 1 sin 2 π 5 = 2 π 5 .

Prove that: 2 tan − 1 1 2 + tan − 1 1 7 = tan − 1 31 17

We have to show that, 2 tan − 1 1 2 + tan − 1 1 7 = tan − 1 31 17 Now, Taking L . H . S . = 2 tan - 1 1 2 + tan - 1 1 7 = tan - 1 2 × 1 2 1 - 1 2 2 + tan - 1 1 7 ∵ 2 tan - 1 x = tan - 1 2 x 1 - x 2 , x ∈ - 1 , 1 = tan - 1 1 1 - 1 4 + tan - 1 1 7 = tan - 1 4 3 + tan - 1 1 7 = tan - 1 4 3 + 1 7 1 - 4 3 × 1 7 ∵ tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y = tan - 1 28 + 3 21 21 - 4 21 = tan - 1 31 17 = R . H . S . ∴ L . H . S = R . H . S Hence proved

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 2 - Inverse Trigonometric Functions>Karnataka Board-2020

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

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

Attempt the free practice questions on Chapter 2: Inverse Trigonometric Functions, Exercise 1: Karnataka 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.