Question

If $\alpha$ & $\beta$ 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

We have,

$a\tan \theta +b\sec \theta =c\to \left(1\right)$

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

$\Rightarrow {(c-a\tan \theta )}^2=b^2{\sec }^2\theta$

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

$\Rightarrow {\tan }^2\theta (a^2-b^2)-2ac\tan \theta -(c^2-b^2)=0\to \left(2\right)$

It is given that $\alpha$ and $\beta$ are the solutions of equation $\left(1\right)$ so, $tan\alpha$ and $tan\beta$ are the roots of $\left(2\right)$.

Hence,

$\tan \alpha +\tan \beta =\frac{2ac}{a^2-b^2}$

and $\tan \alpha .\tan \beta =\frac{(c^2-b^2)}{(a^2-b^2)}$

Now, $\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$

$=\frac{\frac{2ac}{a^2-b^2}}{1-\frac{c^2-b^2}{a^2-b^2}}=\frac{2ac}{a^2-c^2}$

Hence proved