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
 A: 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. 
A: 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,
$$\tan(\alpha+\beta)=\frac{\tan \alpha +\tan\beta}{1-\tan \alpha \tan \beta}$$
$$\tan(\alpha+\beta)=\frac{\frac{2ac}{a^2-b^2}}{1-\frac{c^2-b^2}{a^2-b^2}}$$
$$\tan(\alpha  + \beta)=\frac{2ac}{a^2-c^2}$$
A: I think there is mistake in your computations. 
$$(a\sin\theta+b)^2=c^2(1-\sin^2\theta),$$ 
which does not give, what you wish.
