If \(\sin x-\sin y=\frac{1}{2}\) and \(\cos x-\cos y=1\), then \(\tan (x+y)\) is equal to

Given,
\(\sin x-\sin y=\frac{1}{2}...(i)\)
and \(\cos x-\cos y=1...(ii)\)
\(\Rightarrow \quad 2 \cos \frac{x+y}{2} \cdot \sin \frac{x-y}{2}=\frac{1}{2}...(iii)\)
and \(-2 \sin \frac{x+y}{2} \cdot \sin \frac{x-y}{2}=1...(iv)\)
On dividing Eq. (iv) by Eq. (iii), we get
\(\begin{gathered}
-\tan \left(\frac{x+y}{2}\right)=2 \\
\Rightarrow \quad \tan \left(\frac{x+y}{2}\right)=-2 \ldots(v) \\
\text { Now, } \tan (x+y)=\frac{2 \tan \left(\frac{x+y}{2}\right)}{1-\tan ^{2}\left(\frac{x+y}{2}\right)}
\end{gathered}\)
\(\left(\because \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right)\)
\(=\frac{2(-2)}{1-(-2)^{2}} \quad\) [using Eq. (v)]
\(=\frac{-4}{1-4}=\frac{-4}{-3}=\frac{4}{3}\)