If $\tan \theta +\tan \varphi =a,\cot \theta +\cot \varphi =b,\theta -\varphi =\alpha \ne 0$ then

   $\frac{a}{b}=\tan \theta \tan \varphi$

${\tan }^2\alpha ={\tan }^2(\theta -\varphi )={\left[\frac{\tan \theta -\tan \varphi }{1+\tan \theta \tan \varphi }\right]}^2$

        $\begin{array}{r}=\frac{a^2-4(a/b)}{[1+(a/b){]}^2}\\ =\frac{ab(ab-4)}{(a+b{)}^2}\end{array}$

as ${\tan }^2\alpha >0,ab>4.$