Question

prove that tan pi by 4 + theta by 2 equals to 1 + sin theta upon cos theta equals to cos theta upon 1 minus tan theta​

Answer 1

Answer:

Step-by-step explanation:

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)

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