If ${\cot }^2\mathrm{x}=\cot (\mathrm{x}-\mathrm{y})\cot (\mathrm{x}-\mathrm{z})$ ,then cot2x is equal to (where $\mathrm{x}\ne \pm \mathrm{\pi }/4$) 

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

$\begin{array}{l}\mathrm{or} {\cot }^2\mathrm{x}=\left(\frac{\cot \mathrm{xcot}\mathrm{y}+1}{\cot \mathrm{y}-\cot \mathrm{x}}\right)\left(\frac{\cot \mathrm{xcot}\mathrm{z}+1}{\cot \mathrm{z}-\cot \mathrm{x}}\right)\\ \mathrm{or} {\cot }^2\mathrm{xcot}\mathrm{ycot}\mathrm{z}-{\cot }^3\mathrm{xcot}\mathrm{y}-{\cot }^3\mathrm{xcot}\mathrm{z}+{\cot }^4\mathrm{x}\\ ={\cot }^2\mathrm{xcot}\mathrm{ycot}\mathrm{z}+\cot \mathrm{xcot}\mathrm{y}+\cot \mathrm{xcot}\mathrm{z}+1\\ \mathrm{or} {\cot }^3\mathrm{x}(\cot \mathrm{y}+\cot \mathrm{z})+\cot \mathrm{x}(\cot \mathrm{y}+\cot \mathrm{z})+1-{\cot }^4\mathrm{x}=0\\ \mathrm{or} \cot \mathrm{x}(\cot \mathrm{y}+\cot \mathrm{z})\left(1+{\cot }^2\mathrm{x}\right)+\left(1-{\cot }^2\mathrm{x}\right)\left(1+{\cot }^2\mathrm{x}\right)=0\\ \mathrm{or} \left[\cot \mathrm{x}(\cot \mathrm{y}+\cot \mathrm{z})+\left(1-{\cot }^2\mathrm{x}\right)\right]=0\\ \mathrm{or} \frac{{\cot }^2\mathrm{x}-1}{2\cot \mathrm{x}}=\frac{1}{2}(\cot \mathrm{y}+\cot \mathrm{z})\end{array}$