If $\cos x+\cos y=\frac{4}{5}$ , $\cos x-\cos y=\frac{2}{7}$ , then $14\tan \left(\frac{x-y}{2}\right)+5\cot \left(\frac{x+y}{2}\right)=$

$\cos x+\cos y=\frac{4}{5}$   ---------(1)

$2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x+y}{2}\right)=\frac{4}{5}$

$\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x+y}{2}\right)=\frac{2}{5}$

$\cos x-\cos y=\frac{2}{7}$     -----------(2)

$-2\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)=\frac{2}{7}$

$-\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)=\frac{1}{7}$

$Dividing\left(2\right)by\left(1\right),weget$

$-\frac{\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)}{\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)}=\frac{5}{14}$

$\Rightarrow 14\tan \left(\frac{x+y}{2}\right)=-5\cot \left(\frac{x-y}{2}\right)$

$\Rightarrow 14\tan \left(\frac{x+y}{2}\right)+5\cot \left(\frac{x-y}{2}\right)=0$