Question
$\cot 4 x(\sin 5 x+\sin 3 x)=\cot x(\sin 5 x-\sin 3 x)$

Answer

$\begin{aligned}
\text { L.H.S. } & =\cot 4 x(\sin 5 x+\sin 3 x) \\
& =\cot 4 x\left[2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)\right] \\
& =\frac{\cos 4 x}{\sin 4 x}(2 \sin 4 x \cos x) \\
& =2 \cos 4 x \cos x \\
\text { R.H.S. } & =\cot x(\sin 5 x-\sin 3 x) \\
& =\cot x\left[2 \cos \left(\frac{5 x+3 x}{2}\right) \sin \left(\frac{5 x-3 x}{2}\right)\right] \\
& =\frac{\cos x}{\sin x}(2 \cos 4 x \sin x) \\
& =2 \cos 4 x \cos x
\end{aligned}$
From (i) and (ii), we get L.H.S. $=$ R.H.S.

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