Question

If $\alpha$ and $\beta$ be two distinct roots of the equation $a\tan \theta +b\sec \theta =c$, prove that $\tan (\alpha +\beta )=\left(2ac\right)/(a^2-c^2)$.

Answer 1

Consider the problem

we have,

$\begin{array}{c}a\tan \theta +b\sec \theta =c......\left(1\right)\\ \Rightarrow c-a\tan \theta =b\sec \theta \\ \Rightarrow {\left(c-a\tan \theta \right)}^2={\left(b\sec \theta \right)}^2\\ \Rightarrow c^2-a^2{\tan }^2\theta -2ac\tan \theta =b^2\left(1+{\tan }^2\theta \right)\\ \Rightarrow {\tan }^2\theta \left(a^2-b^2\right)-2ac\tan \theta +\left(c^2-b^2\right)=0......\left(2\right)\end{array}$

It is given that $\alpha$ and $\beta$ are the solutions of the 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,

$\begin{array}{c}\tan \left(\alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha .\tan \beta }\\ =\frac{\frac{2ac}{a^2-b^2}}{1-\frac{c^2-b^2}{a^2-b^2}}\\ =\frac{2ac}{a^2-c^2}\end{array}$

Hence,

Prove that,

$\tan \left(\alpha +\beta \right)=\frac{2ac}{a^2-c^2}$