If $\alpha $ and $\beta$ are roots of equation $a\tan\theta +b \sec\theta=c$. Prove that $\tan(\alpha + \beta)=\frac{2ac}{a^2-c^2}$ If $\alpha $ and $\beta$ are roots of equation $a\tan\theta +b \sec\theta=c$. Prove that $\tan(\alpha  + \beta)=\frac{2ac}{a^2-c^2}$
i have converted tan to sin and cos and reached to $\sin^2\theta(a^2-c^2+2ac) + c^2-b^2-ac=0$. how do i proceed
thanks

We can write the quadratic equation as: $$a\tan \theta + b\sec \theta = c $$ $$\Rightarrow a\tan \theta + b\sqrt {\tan^2 \theta + 1} = c $$ $$\Rightarrow (a^2-b^2)\tan^2 \theta -2ac \tan \theta +(c^2-b^2)=0$$
Now $$\tan (\alpha +\beta ) =\frac {\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}. $$ Can you take it from here? Hope it helps. 
