We know that,
sinA-sinB=2cos\(( \dfrac{A+B}{2})sin( \dfrac{A-B}{2})\),
cosA+cosB=2cos\( ( \dfrac{A+B}{2})cos( \dfrac{A-B}{2})\)
\( \dfrac{sinx-siny}{cosx+cosy}=tan\dfrac{x-y}{2}\)
L.H.S.\( \dfrac{sinx-siny}{cosx+cosy}\)
= \(\dfrac{2cos(\dfrac{x+y}{2}).sin(\dfrac{x-y}{2})}{2cos(\dfrac{x+y}{2}).cos(\dfrac{x-y}{2})}\)
= \( \dfrac{sin(\dfrac{x-y}{2})}{cos(\dfrac{x-y}{2})}\)
= \( tan(\dfrac{x-y}{2})\)
L.H.S=R.H.S
Hence proved.
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