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Prove that
$\sin ^2 6 x-\sin ^2 4 x=\sin 2 x \sin 10 x$
$\sin ^2 6 x-\sin ^2 4 x=\sin 2 x \sin 10 x$
Solution:
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Verified Answer
$\begin{aligned} \text { L.H.S } &=\sin ^2 6 x-\sin ^2 4 x \\ &=\frac{1-\cos 12 x}{2}-\frac{1-\cos 8 x}{2} \\ &=\frac{\cos 8 x-\cos 12 x}{2} \\ &=\frac{2 \sin 10 x \cdot \sin 2 x}{2} \\ &=\sin 10 x \cdot \sin 2 x=\text { RHS. } \end{aligned}$
Proved.
Proved.
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