Question

1/1+cosθ +1/1-cosθ=2cosec²θ ,Prove it by using trigonometric identities.

Answer 1

LHS = 1/(1 + cosθ) + 1/(1 - cosθ)

= {(1 - cosθ) + (1 + cosθ)}/(1 + cosθ)(1 - cosθ)

use formula , (a - b)(a + b) = a² - b² then, (1 - cosθ)(1 + cosθ) = 1 - cos²θ ----(1)

= 2/(1 - cos²θ) [ from equation (1), ]

we know, sin²x + cos²x = 1 so, 1 - cos²θ = sin²θ -----(2)

= 2/sin²θ [ from equation (2), ]

= 2cosec²θ = RHS

Answer 2

HELLO DEAR,

LHS = 1/(1 + cosθ) + 1/(1 - cosθ)

= [(1 - cosθ) + (1 + cosθ)]/(1 + cosθ)(1 - cosθ)

we know:- (a - b)(a + b) = a² - b²

= 2/(1 - cos²θ)

we know, sin²x + cos²x = 1

= 2/sin²θ

= 2cosec²θ = RHS

Hence, L.H.S = R.H.S

I HOPE ITS HELP YOU DEAR,
THANKS