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Question

sin3θ+2sin5θ+sin7θsinθ+2sin3θ+sin5θ=cos2θ+sin2θ.cot3θ.

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Solution

sin3θ+2sinθ+sin7θsinθ+2sin3θ+sin5θ

sin3θ+sin7θ+2sin5θsinθ+sin5θ+2sin3θ

=2sin5θcos2θ+2sin5θ2sin3θcos2θ+2sin3θ.....(sinA+sinB=2sin(A+B)2cos(AB)2)

=2sin5θ(cos2θ+1)2sin3θ(cos2θ+1)

=sin5θsin3θLHS

Now, RHS , cos2θ+sin2θcot3θ

=cos2θ+sin2θ×cos3θsin3θ

sin3θcos2θ+sin2θcos3θsin3θ

=sin5θsin3θ......(sin(A+B)=sinAcosB+cosAsinB)

=LHS

Hence LHS=RHS

Hence proved.

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