Question

(2-cos theta) whole square is equals to
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Answer 1

Answer:

LHS=(1−sinθ+cosθ)

2

LHS=1+sin

2

θ+cos

2

θ−2sinθ+2cosθ−2sinθcosθ

LHS=2−2sinθ+2cosθ−2sinθcosθ

LHS=2(1−sinθ)+2cosθ(1−sinθ)

LHS=2(1−sinθ)(1+cosθ)=RHS

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