prove that 1+sintheta-costheta/1+sintheta+costheta=tan theta/2
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LHS of the given equation is:
1+sinθ−cosθ1+sinθ+cosθ=1+sinθ−(1−2sin2θ2)1+sinθ+2cos2θ2−1 [since cos2θ=2cos2θ2−1=1−2sin2θ2=2sinθ2.cosθ2+2sin2θ22sinθ2.cosθ2+2cos2θ2 [since sin2θ=2sinθ.cosθ=2sinθ2.(cosθ2+sinθ2)2cosθ2.(cosθ2+sinθ2)=sinθ2cosθ2=tanθ2
= RHS
hope this helps you
1+sinθ−cosθ1+sinθ+cosθ=1+sinθ−(1−2sin2θ2)1+sinθ+2cos2θ2−1 [since cos2θ=2cos2θ2−1=1−2sin2θ2=2sinθ2.cosθ2+2sin2θ22sinθ2.cosθ2+2cos2θ2 [since sin2θ=2sinθ.cosθ=2sinθ2.(cosθ2+sinθ2)2cosθ2.(cosθ2+sinθ2)=sinθ2cosθ2=tanθ2
= RHS
hope this helps you
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