Question

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

Answer 1

Here $b\sec \theta =c-a\tan \theta$. Square
$b^2(1+{\tan }^2\theta )=c^2-2ca\tan \theta +a^2{\tan }^2\theta$
$(a^2-b^2){\tan }^2\theta -2ca\tan \theta +(c^2-a^2)=0$
$\therefore \tan \alpha +\tan \beta =\frac{2ca}{a^2-b^2},\tan \alpha \tan \beta =\frac{c^2-a^2}{a^2-b^2}$
$\therefore \tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\frac{2ca}{(a^2-b^2)-(c^2-a^2)}=\frac{2ac}{a^2-c^2}$