Question

prove that sin theta divided by 1+cos theta. + 1+cos theta divided by sin theta =2 cosec theta​

Answer 1

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