If cosec θ + cot θ = P , show that P 2 + 1 P 2 - 1 = sec θ .

Given, cosec θ + cot θ = P ------- i We know that cosec 2 θ - cot 2 θ = 1 . ⇒ cosec θ - cot θ cosec θ + cot θ = 1 ⇒ P cosec θ - cot θ = 1 ⇒ cos e c θ - cot θ = 1 P ------ i i Adding i and ii , 2 cos e c θ = P + 1 P ⇒ 2 cos e c θ = P 2 + 1 P ⇒ cos e c θ = P 2 + 1 2 P ⇒ sin θ = 1 cosec θ = 2 P P 2 + 1 Using the identity: sin 2 θ + cos 2 θ = 1 ⇒ cos θ = 1 - 2 P P 2 + 1 2 ⇒ cos θ = ( P 2 + 1 ) 2 - 4 P 2 P 2 + 1 2 ⇒ cos θ = P 4 + 1 + 2 P 2 - 4 P 2 P 2 + 1 2 ⇒ cos θ = P 2 - 1 2 P 2 + 1 2 ⇒ cos θ = P 2 - 1 P 2 + 1 ∴ sec θ = 1 cos θ = P 2 + 1 P 2 - 1 Hence, proved.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 11 - Trigonometry>Andhra Pradesh Board-2019

Embibe Experts Solutions for Chapter: Trigonometry, Exercise 1: Andhra Pradesh Board-2019

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Trigonometry, Exercise 1: Andhra Pradesh Board-2019

Attempt the free practice questions on Chapter 11: Trigonometry, Exercise 1: Andhra Pradesh 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.