1/1+cosθ +1/1-cosθ=2cosec²θ ,Prove it by using trigonometric identities.
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Answered by
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
= {(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
Answered by
3
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
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