Question
$\sin ^2 6 x-\sin ^2 4 x=\sin 2 x \sin 10 x$
✓
Answer
$\begin{aligned}
\text { L.H.S. } & =\sin ^2 6 x-\sin ^2 4 x \\
& =(\sin 6 x)^2-(\sin 4 x)^2 \\
& =(\sin 6 x+\sin 4 x)(\sin 6 x-\sin 4 x) \\
& =\left[2 \sin \left(\frac{6 x+4 x}{2}\right) \cos \left(\frac{6 x-4 x}{2}\right)\right] \\
& {\left[2 \cos \left(\frac{6 x+4 x}{2}\right) \sin \left(\frac{6 x-4 x}{2}\right)\right] } \\
& =(2 \sin 5 x \cos x)(2 \cos 5 x \sin x) \\
& =(2 \sin x \cos x)(2 \sin 5 x \cos 5 x) \\
& =\sin 2 x \sin 10 x \\
& =\text { R.H.S.}
\end{aligned}$
\text { L.H.S. } & =\sin ^2 6 x-\sin ^2 4 x \\
& =(\sin 6 x)^2-(\sin 4 x)^2 \\
& =(\sin 6 x+\sin 4 x)(\sin 6 x-\sin 4 x) \\
& =\left[2 \sin \left(\frac{6 x+4 x}{2}\right) \cos \left(\frac{6 x-4 x}{2}\right)\right] \\
& {\left[2 \cos \left(\frac{6 x+4 x}{2}\right) \sin \left(\frac{6 x-4 x}{2}\right)\right] } \\
& =(2 \sin 5 x \cos x)(2 \cos 5 x \sin x) \\
& =(2 \sin x \cos x)(2 \sin 5 x \cos 5 x) \\
& =\sin 2 x \sin 10 x \\
& =\text { R.H.S.}
\end{aligned}$
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