prove that sin theta divided by 1+cos theta. + 1+cos theta divided by sin theta =2 cosec theta
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Step-by-step explanation:
Given:-
[Sin θ/(1+Cos θ)] + [(1+ Cos θ)/Sin θ]
To find:-
Prove that
[Sin θ/(1+Cos θ)] + [(1+ Cos θ)/Sin θ] = 2 Cosec θ
Solution:-
LHS:-
[Sin θ/(1+Cos θ)] + [(1+ Cos θ)/Sin θ]
=>[( Sin θ× Sin θ)+(1+Cosθ)(1+Cosθ)]/Sinθ(1+Cos θ)
=> [Sin^2 θ+(1+Cos θ)(1+Cosθ)]/Sinθ(1+Cos θ)
We know that Sin^2 A + Cos^2 A = 1
=> Sin^2 A = 1 - Cos^2 A
=> [(1- Cos^2 θ)+(1+Cos θ)(1+Cosθ)]/Sinθ(1+Cos θ)
(1-Cosθ)(1+Cosθ)+(1+Cos θ)(1+Cosθ)]/Sinθ(1+Cos θ)
=> (1+Cosθ)[(1-Cosθ)+(1+Cosθ)]/Sinθ(1+Cos θ)
On Cancelling (1+ Cos θ) in both numerator and denominator
=> [(1-Cosθ)+(1+Cosθ)]/Sinθ
=> [1-Cosθ+1+Cosθ]/Sinθ
=> (1+1)/ Sin θ
=> 2/ Sin θ
=> 2(1/ Sin θ)
=> 2 Cosec θ
=> RHS
LHS = RHS
Hence Proved
Answer:-
[Sin θ/(1+Cos θ)] + [(1+ Cos θ)/Sin θ] = 2 Cosec θ
Used formulae:-
- Sin^2 A + Cos^2 A = 1
- Sin^2 A = 1 - Cos^2 A
- 1/Sin A = Cosec A
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