If $\frac{\cos ({\theta }_1+{\theta }_2)}{\cos ({\theta }_1-{\theta }_2)}+\frac{\cos ({\theta }_3-{\theta }_4)}{\cos ({\theta }_3+{\theta }_4)}=0,$ then $\cot {\theta }_1\cdot \cot {\theta }_2\cdot \cot {\theta }_3\cdot \cot {\theta }_4=$

$\frac{\cos ({\theta }_1+{\theta }_2)}{\cos ({\theta }_1-{\theta }_2)}=\frac{\cos ({\theta }_3-{\theta }_4)}{-\cos ({\theta }_3+{\theta }_4)}$

Using componendo and dividendo and transformation

$\frac{\cos ({\theta }_1+{\theta }_2)+\cos ({\theta }_1-{\theta }_2)}{\cos ({\theta }_1+{\theta }_2)-\cos ({\theta }_1-{\theta }_2)}=\frac{\cos ({\theta }_3-{\theta }_4)-\cos ({\theta }_3+{\theta }_4)}{\cos ({\theta }_3-{\theta }_4)+\cos ({\theta }_3+{\theta }_4)}$

$\frac{2\cos {\theta }_1\cos {\theta }_2}{-2\sin {\theta }_1\sin {\theta }_2}=\frac{2\sin {\theta }_3\sin {\theta }_4}{2\cos {\theta }_3\cos {\theta }_4}$

$\frac{-1}{\tan {\theta }_1\tan {\theta }_2}=\tan {\theta }_3\tan {\theta }_4$

$\tan {\theta }_1\tan {\theta }_2\tan {\theta }_3\tan {\theta }_4=-1$