If sin − 1 ⁡ x = tan − 1 ⁡ y , then show that, 1 x 2 − 1 y 2 = 1 .

Let sin − 1 ⁡ x = tan − 1 ⁡ y = θ ⇒ x = sin ⁡ θ and y = tan ⁡ θ Now, 1 x 2 = 1 sin 2 ⁡ θ = cosec 2 ⁡ θ and 1 y 2 = 1 tan 2 ⁡ θ = cot 2 ⁡ θ Consider, 1 x 2 − 1 y 2 = cosec 2 ⁡ θ − cot 2 ⁡ θ = 1 ∵ 1 + cot 2 A = cosec 2 A Hence, proved.

Prove that tan π 4 + 1 2 cos - 1 a b + tan π 4 - 1 2 cos - 1 a b = 2 b a .

To prove: tan π 4 + 1 2 cos - 1 a b + tan π 4 - 1 2 cos - 1 a b = 2 b a Consider L . H . S . = tan π 4 + 1 2 cos - 1 a b + tan π 4 - 1 2 cos - 1 a b Put 1 2 cos - 1 a b = θ ⇒ cos 2 θ = a b . . . . . i Therefore, tan π 4 + 1 2 cos - 1 a b + tan π 4 - 1 2 cos - 1 a b = tan π 4 + θ + tan π 4 - θ = tan π 4 + tan θ 1 - tan π 4 × tan θ + tan π 4 - tan θ 1 + tan π 4 × tan θ = 1 + tan θ 1 - tan θ + 1 - tan θ 1 + tan θ = 1 + tan θ 2 + 1 - tan θ 2 1 - tan 2 θ = 1 + tan 2 θ + 2 tan θ + 1 + tan 2 θ - 2 tan θ 1 - tan 2 θ = 2 + 2 tan 2 θ 1 - tan 2 θ = 2 1 + tan 2 θ 1 - tan 2 θ = 2 cos 2 θ [Using cos 2 x = 1 - tan 2 x 1 + tan 2 x ] = 2 a b [Using i equation] = 2 b a = R . H . S . Hence, proved.

If tan - 1 x + tan - 1 y + tan - 1 z = π 2 and x + y + z = 3 , then show that x = y = z .

Given that, tan - 1 x + tan - 1 y + tan - 1 z = π 2 and x + y + z = 3 . We know that, tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y ⇒ tan - 1 x + y 1 - x y + tan - 1 z = π 2 ⇒ tan tan - 1 x + y 1 - x y = tan π 2 - tan - 1 z ⇒ tan - 1 x + y 1 - x y = cot - 1 z We know that, cot - 1 x = tan - 1 1 x ⇒ tan - 1 x + y 1 - x y = tan - 1 1 z Now, applying tan on both sides ⇒ tan tan - 1 x + y 1 - x y = tan tan - 1 1 z We know that, tan tan - 1 x = x ⇒ x + y 1 - x y = 1 z ⇒ x z + y z = 1 - x y ⇒ x y + x z + y z = 1 .....(i) Given equation is x + y + 2 = 3 Squaring on both sides, ( x + y + z ) 2 = 3 ⇒ x 2 + y 2 + z 2 + 2 ( x y + y z + z x ) = 3 From equation(i) ⇒ x 2 + y 2 + z 2 + 2 1 = 3 ⇒ x 2 + y 2 + z 2 = 1 ⇒ x 2 + y 2 + z 2 - 1 = 0 .....(ii) Put equation (i) in (ii) x 2 + y 2 + z 2 - ( x y + y z + z x ) = 0 ⇒ 2 x 2 + y 2 + z 2 - 2 ( x y + y z + z x ) = 0 ⇒ x 2 - 2 x y + y 2 + y 2 - 2 y z + z 2 + z 2 - 2 z x + x 2 = 0 ⇒ ( x - y ) 2 + ( y - z ) 2 + ( z - x ) 2 = 0 Here, x - y = 0 , y - z = 0 and z - x = 0 ⇒ x = y , y = z , z = x ⇒ x = y = z Hence, showed.

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Relations and Functions>Chapter 2 - Inverse Trigonometric Functions>West Bengal Board-2019

Embibe Experts Solutions for Exercise 1: West Bengal Board-2019

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Exercise 1: West Bengal Board-2019

Attempt the practice questions from Exercise 1: West Bengal Board-2019 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.