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Prove that,
$\cot 4 x[\sin 5 x+\sin 3 x]=\cot x(\sin 5 x-\sin 3 x)$
$\cot 4 x[\sin 5 x+\sin 3 x]=\cot x(\sin 5 x-\sin 3 x)$
Solution:
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Verified Answer
$\begin{aligned} \text { LHS } &=\cot 4 x[\sin 5 x+\sin 3 x] \\ &=\cot 4 x[2 \sin 4 x \cos x] \end{aligned}$
$=\frac{\cos 4 x}{\sin 4 x} \times 2 \sin 4 x \cos x=2 \cos x \cos 4 x$
$=\frac{\cos x}{\sin x} \times(2 \cos 4 x \sin x)$
$=\frac{\cos x}{\sin x} \times(\sin 5 x-\sin 3 x)$
${[\because 2 \cos \mathrm{A} \sin \mathrm{B}=\sin (\mathrm{A}+\mathrm{B})-\sin (\mathrm{A}-\mathrm{B})] }$
$=\cot x(\sin 5 x-\sin 3 x)=\mathrm{RHS}$
$=\frac{\cos 4 x}{\sin 4 x} \times 2 \sin 4 x \cos x=2 \cos x \cos 4 x$
$=\frac{\cos x}{\sin x} \times(2 \cos 4 x \sin x)$
$=\frac{\cos x}{\sin x} \times(\sin 5 x-\sin 3 x)$
${[\because 2 \cos \mathrm{A} \sin \mathrm{B}=\sin (\mathrm{A}+\mathrm{B})-\sin (\mathrm{A}-\mathrm{B})] }$
$=\cot x(\sin 5 x-\sin 3 x)=\mathrm{RHS}$
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