If ${\cot }^{2}x=\cot (x-y)\cdot \cot (x-z)$, where $x\ne \pm \frac{\pi }{4}$, then $\cot 2x$ is equal to

${\cot }^{2}x=\cot (x-y)\cdot \cot (x-z)$

$\frac{{\cos }^{2}x}{{\sin }^{2}x}=\frac{\cos (x-y)\cdot \cos (x-z)}{\sin (x-y)\cdot \sin (x-z)}$

$\frac{{\cos }^{2}x-{\sin }^{2}x}{{\cos }^{2}x+{\sin }^{2}x}=\frac{\cos (x-y)\cos (x-z)-\sin (x-y)\sin (x-z)}{\cos (x-y)\cos (x-z)+\sin (x-y)\sin (x-z)}$

$\Rightarrow \cos 2x=\frac{\cos [\overline{x-y}+\overline{x-z}]}{\cos [\overline{x-y}-\overline{x-z}]}$

$\Rightarrow \cos 2x\cdot \cos (y-z)=\cos (2x-(y+z))$

$\Rightarrow \cos 2x\cdot \cos (y-z)=\cos (2x-(y+z))$

$=\cos 2x\cdot \cos (y+z)+\sin 2x\cdot \sin (y+z)$

$\Rightarrow \cos 2x[2\sin y\cdot \sin z]=\sin 2x\cdot \sin (y+z)$

$\Rightarrow \cot 2x=\frac{1}{2}\cdot \frac{\sin (y+z)}{\sin y\cdot \sin z}$

$\Rightarrow \cot 2x=\frac{1}{2}(\cot y+\cot z)$.