Math, asked by pravdeep3333, 11 months ago

2-cosec2A/cosec2A+2cot A=sinA-cosA/sinA+cosA

Answers

Answered by saumik61
5
LHS
( 2 - cosec²θ)/(cosec²θ+ 2cot θ)
= (2-1-cot²θ)(1+cot²θ+2cotθ)
=(1-cot²θ)/(1+cotθ)²
=(1-cotθ)(1+cotθ) /(1+cotθ)²
=(1-cotθ)/(1+cotθ)
={1-(cosθ/sinθ)}/{1+(cosθ/sinθ)}
{(sinθ-cosθ)/sinθ/{(sinθ+cosθ)/sinθ}
=(sinθ-cosθ)/(sinθ+cosθ)=RHS
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