<s>[INST] 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
 [/INST] We have $a\tan\theta +b \sec\theta=c$. So,
\begin{align*}
c-a\tan \theta & = b \sec \theta\\
(c-a\tan \theta)^2 & = (b\sec \theta)^2\\
 (a^2-b^2)\tan^2 \theta -2ac \tan \theta+(c^2-b^2) & = 0.
\end{align*}
This is a quadratic in $\tan \theta$. It has two solutions $\tan \alpha$ and $\tan \beta$. So 
\begin{align*}
 \tan \alpha + \tan \beta & = \frac{2ac}{a^2-b^2}\\
 \tan \alpha \cdot \tan \beta & = \frac{c^2-b^2}{a^2-b^2}
\end{align*}
Now,
$$\tan(\alpha+\beta)=\frac{\tan \alpha +\tan\beta}{1-\tan \alpha \tan \beta}$$
$$\tan(\alpha+\beta)=\frac{\frac{2ac}{a^2-b^2}}{1-\frac{c^2-b^2}{a^2-b^2}}$$
$$\tan(\alpha  + \beta)=\frac{2ac}{a^2-c^2}$$
</s>