Ta có $\tan \alpha+ \cot \beta=\frac{\sin \alpha}{\cos \alpha}+\frac{\cos \beta}{\sin \beta}=\frac{\cos (\alpha- \beta)}{\cos \alpha.\sin \beta}=\frac{2\cos( \alpha - \beta)}{\sin( \beta + \alpha)+\sin(\beta - \alpha)}$Áp dụng với $\alpha=\frac{x+y}{2}, \beta = \frac{x-y}{2}$
Ta có $\tan \frac{x+y}{2}+\cot \frac{x-y}{2}=\frac{2\cos y}{\sin x-\sin y}$
$\Rightarrow VT=\frac{\sin^2x-\sin^2y}{2\cos y}$
Nên ta chỉ cần cm $\sin^2x-\sin^2y=\sin(x+y).\sin(x-y)$
$\Leftrightarrow (1-\cos^2x)+(-1+\cos^2y)=\frac{-1}2\left[\cos2x -\cos2y\right]$
$\Leftrightarrow \cos^2y-\cos^2x= \frac{\cos 2y+1}2-\frac{\cos 2x+1}{2}$ (luôn đúng)