$\alpha$ and $\beta$ are solution of $a \cdot \tan\theta + b \cdot \sec\theta = c$ show $ \tan(\alpha + \beta) = \frac{2ac}{a^2 - c^2}$ If $\alpha$ and $\beta$ are the solution of $$a \cdot \tan\theta + b \cdot \sec\theta = c$$, then show that $$ \tan(\alpha + \beta) = \frac{2ac}{a^2 - c^2}$$
I did the following:

$$a \cdot \tan\theta + b \cdot \sec\theta = c$$ 
Squaring both sides,  $$a^2 \cdot \tan^2\theta + b^2 \cdot \sec \theta + 2 \cdot a \cdot b \cdot \tan \theta \cdot \sec \theta = c^2$$
This implies $$(a^2 + b^2) \cdot \sin^2 \theta + 2 \cdot a \cdot b \cdot \sin \theta + (b^2 - c^2) = 0$$ 
Hence I get $$\sin \alpha + \sin \beta = \frac{-2ab}{a^2 + c^2}$$ and $$\sin \alpha \cdot \sin \beta = \frac{b^2 - c^2}{a^2 - c^2}$$

But I am not able to solve it further.

Consider the following:
\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*}
Think of this as 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 you can compute using
$$\tan(\alpha+\beta)=\frac{\tan \alpha +\tan\beta}{1-\tan \alpha \tan \beta}$$
