Question

If $\sin x+\sin y=a$ and $\cos x+\cos y=b$ show that
$\tan \left(\frac{x-y}{2}\right)=\pm \sqrt{\left(\frac{4-a^2-b^2}{a^2+b^2}\right)}$

Answer 1

$\because {\tan }^2\left(\frac{A}{2}\right)=\frac{1-casA}{1+\cos A}$
$\therefore {\tan }^2\left(\frac{x-y}{2}\right)=\frac{1-(a^2+b^2-2)/2}{1+(a^2+b^2-2)/2}$

or $\tan$ $\left(\frac{x-y}{2}\right)=\pm \sqrt{\left(\frac{4-a^2-b^2}{a^2+b^2}\right)}$