Question

Prove: cos theta/1+sin theta=tan(pi/4-theta/2)

Answer 1

LHS
=cosθ/1+sinθ
=cosθ(1-sinθ)/(1+sinθ)(1-sinθ)
=cosθ(1-sinθ)/(1²-sin²θ)
=cosθ(1-sinθ)/cos²θ
=(1-sinθ)/cosθ
=[cos²(θ/2)-2sin(θ/2)cos(θ/2)+sin²(θ/2)]/[cos²(θ/2)-sin²(θ/2)]
[∵, sin²(θ/2)+cos²(θ/2)=1, sinθ=2sin(θ/2)cos(θ/2), cosθ=cos²(θ/2)-sin²(θ/2)]
=[cos(θ/2)-sin(θ/2)]²/[{cos(θ/2)+sin(θ/2)}{cos(θ/2)-sin(θ/2)}]
=[cos(θ/2-sin(θ/2)]/[cos(θ/2)+sin(θ/2)]
RHS
=tan(π/4-θ/2)
=[tan(π/4)-tan(θ/2)]/[1+tan(π/4)tan(θ/2)]
=[1-tan(θ/2)]/[1+tan(θ/2)]
=[1-sin(θ/2)/cos(θ/2)]/[1+sin(θ/2)/cos(θ/2)]
=[{cos(θ/2)-sin(θ/2)}/cos(θ/2)]/[{cos(θ/2)+sin(θ/2)}/cos(θ/2)]
=[cos(θ/2)-sin(θ/2)]/[cos(θ/2)+sin(θ/2)]
∴, LHS=RHS (Proved)

Answer 2

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