2015/064) If $\alpha$ and $\beta$ are the solutions of the equation $a \tan \theta + b \sec \theta = c$ , then show that $\tan (\alpha + \beta)= 2\frac{ac}{a^2-c^2}$

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

so $b^2 \sec^2 \theta = (c-a \tan \theta)^2$

or $b^2 ( 1 + \tan ^2 \theta) = c^2 + a^2 \tan ^2 \theta - 2ac \tan \theta$

or $(a^2-b^2) \tan ^2 \theta - 2ac \tan \theta + (c^2-b^2) = 0$

or $tan ^2 \theta - \dfrac{2ac}{a^2-b^2} + \dfrac{c^2-b^2}{a^2-b^2} = 0 \cdots(1)$

if $\alpha$ and $\beta$ are the solutions of the equation $a \tan \theta + b \sec \theta = c$

then $\tan \alpha$ and $\tan \beta$ are the solutions of (1)

so $tan\, \alpha + \tan\, \beta = \dfrac{2ac}{a^2-b^2}\cdots (2)$

$tan\, \alpha \cdot \tan\, \beta = \dfrac{c^2-b^2}{a^2-b^2}\cdots (3)$ 



so $\tan (\alpha  + \beta) = \dfrac{ \tan\,\alpha + \tan\, \beta}{1- \tan\, \alpha \tan\,\beta}$
 = $\dfrac{\frac{2ac}{a^2-b^2}}{1- \frac{c^2-b^2}{a^2-b^2}}$
= $\dfrac{2ac}{a^2-c^2}$

PROVED