We know that,
sinA+sinB=2sin\( ( \dfrac{A+B}{2})cos( \dfrac{A-B}{2})\)
And
sinA-sinB=2cos\(\)\( ( \dfrac{A+B}{2})sin( \dfrac{A-B}{2})\)
L.H.S.=sin26x-sin24xa
=[sin6x+sin4x)(sin6x-sin4x)
\( [ (2sin \dfrac{6x+4x}{2})cos( \dfrac{6x-4x}{2})]\)
\( [ (2cos \dfrac{6x+4x}{2}).sin( \dfrac{6x-4x}{2})]\)
=(2sin5xcosx)(2cos5xsinx)
Now we know that, sin2x=2sinxcosx ,
Therefore we have,
L.H.S=(2sin5xcos5)(2sinxcosx)
=sin10xsin2x
= R.H.S.
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