=sin26x−sin24x=(sin6x+sin4x)(sin6x−sin4x)=[2sin(6x+4x2).cos(6x−4x2)]×[2cos(6x+4x2)×sin(6x−4x2)]=(2sin5x.cosx)(2cos5x.sinx)=(2sin5x.cos5x)(2sinx.cosx)=sin10x×sin2x

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$\sin^{2} 6 x - \sin^{2} 4 x =$

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$\cos 2 x$

$\sin 2 x$

$\sin 10 x . \sin 2 x$

$\tan 3 x$

answer is C.

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Detailed Solution

$\begin{array}{l} = \sin^{2} 6 x - \sin^{2} 4 x \\ = (\sin 6 x + \sin 4 x) (\sin 6 x - \sin 4 x) \\ = [2 \sin (\frac{6 x + 4 x}{2}) . \cos (\frac{6 x - 4 x}{2})] \times [2 \cos (\frac{6 x + 4 x}{2}) \times \sin (\frac{6 x - 4 x}{2})] \\ = (2 \sin 5 x . \cos x) (2 \cos 5 x . \sin x) \\ = (2 \sin 5 x . \cos 5 x) (2 \sin x . \cos x) \\ = \sin 10 x \times \sin 2 x \end{array}$