We know that
sinA+sinB=2sin\( ( \dfrac{A+B}{2})cos( \dfrac{A-B}{2})\)
cosA+cosB=2cos\( ( \dfrac{A+B}{2})cos( \dfrac{A-B}{2})\)
Now , L.H.S. \( \dfrac{sinx+sin3x}{cosx+cos3x}\)
= \( \dfrac{2sin(\dfrac{x+3x}{2})cos(\dfrac{x-3x}{2})}{2cos(\dfrac{x+3x}{2})cos(\dfrac{x-3x}{2})}\)
= \( \dfrac{sin2x}{cos2x}=tan2x\)
Therefore L.H.S = R.H.S.
Hence proved
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