$\alpha$ 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.

Using Double-Angle Formula,
$$\sec\theta=\dfrac{1+t^2}{1-t^2},\tan\theta=\dfrac{2t}{1-t^2}$$  where $t=\tan\dfrac\theta2$ to form a Quadratic Equation in $t$ whose roots will be $$\tan\dfrac\alpha2,\tan\dfrac\beta2$$
Can you find $$\tan\left(\dfrac\alpha2+\dfrac\beta2\right)$$  and subsequently, $$\tan(\alpha+\beta)$$
