QPaperGen

We use the identities:


\[
\sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right)
\]




\[
\cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)
\]



Now, the left-hand side (LHS) becomes:


\[
\frac{\sin x - \sin y}{\cos x + \cos y}
= \frac{2 \cos\left(\frac{x + y}{2}\right) \sin\left(\frac{x - y}{2}\right)}{2 \cos\left(\frac{x + y}{2}\right) \cos\left(\frac{x - y}{2}\right)}
\]





\[
= \frac{\sin\left(\frac{x - y}{2}\right)}{\cos\left(\frac{x - y}{2}\right)} = \tan\left(\frac{x - y}{2}\right)
\]



Hence, LHS = RHS. (Proved)