prove that $\tan(\alpha+\beta) = 2ac/(a^2-c^2)$ [duplicate]

HINT:

$$b\sec\theta=c-a\tan\theta$$

Squaring we get $$(c-a\tan\theta)^2=(b\sec\theta)^2=b^2(1+\tan^2\theta)$$

Rearrange to form a quadratic equation in $\tan\theta$ whose roots will be $\tan\alpha,\tan\beta$

Now Vieta's formula and expand $\tan(\alpha+\beta)$