Question:
If $\cos x+\cos y=\frac{1}{3}$ and $\sin x+\sin y=\frac{1}{4}$, prove that $\tan \left(\frac{x+y}{2}\right)=\frac{3}{4}$
Solution:
$\cos x+\cos y=\frac{1}{3}$ .............$-i$
$\sin x+\sin y=\frac{1}{4}$ ................-ii
dividing ii by I we get,
$\Rightarrow \frac{\sin x+\sin y}{\cos x+\cos y}=\frac{\frac{1}{4}}{\frac{1}{3}}$
$\Rightarrow \frac{\sin x+\sin y}{\cos x+\cos y}=\frac{3}{4}$
$\Rightarrow \frac{2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}}{2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}}=\frac{3}{4}$
$\Rightarrow \tan \left(\frac{x+y}{2}\right)=\frac{3}{4}$
Using the formula,
$\sin A+\sin B=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$
$\cos A+\cos B=2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$