$\begin{aligned} & \text { We have, } \mathrm{a} \cos 2 \theta+\mathrm{b} \sin 2 \theta=\mathrm{c} \\ & \Rightarrow \mathrm{a}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)+\mathrm{b}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)=\mathrm{c} \\ & \Rightarrow \mathrm{a}-\mathrm{a} \tan ^2 \theta+2 \mathrm{~b} \tan \theta=\mathrm{c}+\mathrm{c} \tan ^2 \theta \\ & \Rightarrow-(\mathrm{a}+\mathrm{c}) \tan ^2 \theta+2 \mathrm{~b} \tan \theta+(\mathrm{a}-\mathrm{c})=0 \\ & \therefore \quad \tan \alpha+\tan \beta=-\frac{2 \mathrm{~b}}{-(\mathrm{c}+\mathrm{a})}=\frac{2 \mathrm{~b}}{\mathrm{c}+\mathrm{a}}\end{aligned}$