If $\alpha ,\beta$ are solutions of $a\cos 2\theta +b\sin 2\theta =c$ , then $\tan \alpha \tan \beta =$

$\begin{array}{l}a\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)+b\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)=c\\ a(1-{\tan }^2\theta )+2b\tan \theta -c(1+{\tan }^2\theta )=0\\ (a+c){\tan }^2\theta -2b\tan \theta +c-a=0\end{array}$

 $\tan \alpha ,\tan \beta$ are the root of the equation

$\tan \alpha .\tan \beta =\frac{c-a}{c+a}$