We know that
sinA+sinB=2sin \( ( \dfrac{A+B}{2})cos( \dfrac{A-B}{2})\),
cosA+cosB=2cos \( ( \dfrac{A+B}{2})cos( \dfrac{A-B}{2})\)
\( \dfrac{sin5x+sin3x}{cos5x+cos3x}=tan4x\)
Now ,
\( L.H.S.= \dfrac{sin5x+sin3x}{cos5x+cos3x}\)
= \( \dfrac{2sin(\dfrac{5x+3x}{2})cos(\dfrac{5x-3x}{2})}{2cos(\dfrac{5x+3x}{2}).cos(\dfrac{5x-3x}{2})}\)
= \( \dfrac{2sin4xcosx}{2cos4xcosx}\)
Further computing we have,
L.H.S=tan4x = R.H.S.
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