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Question

Prove that
2sinθsin2θ2sinθ+sinθ=12tan(θ2)tanθ

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Solution

Here, 2sinθsin2θ2sinθ+sin2θ=2sinθ2sinθcosθ2sinθ+2sinθcosθ [sin2θ=2sinθcosθ]
=2sinθ(1cosθ)2sinθ(1+cosθ)
=1cosθ1+cosθ
=2sin2θ/22cos2θ/2 [cos2θ=12sin2θcos2θ=2cos2θ1]
=tan2θ/2
=12tanθ/2tanθ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢tan2θ=2tanθ1tan2θ1tan2θ=2tanθtan2θtan2θ=12tanθtan2θ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥.

1190055_1327706_ans_90e4f508a5094e059c6d8953ad61f4ab.JPG

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