If $\sin x+\cos y=a$ and $\cos x+\sin y=b$, then $\tan \frac{x-y}{2}$ is equal to

From the given relations we have

$\sin x+\sin \left(\right(\pi /2)-y)=a$ and $\cos x+\cos \left(\right(\pi /2)-y)=b$

$\Rightarrow 2\sin \frac{x+(\pi /2)-y}{2}\cos \frac{x-(\pi /2)+y}{2}=a$

and $2\cos \frac{x+(\pi /2)-y}{2}\cos \frac{x-(\pi /2)+y}{2}=b$

Dividing we get,

$\tan \left(\frac{\pi }{4}+\frac{x-y}{2}\right)=\frac{a}{b}\Rightarrow \frac{1+\tan \frac{x-y}{2}}{1-\tan \frac{x-y}{2}}=\frac{a}{b}$

or  $\tan \frac{x-y}{2}=\frac{a-b}{a+b}$.