Question

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

Answer 1

We have ,
$a\tan \theta +b\sec \theta =c\Rightarrow b\sec \theta =c-a\tan \theta \Rightarrow b^2{\sec }^2\theta =(c-a\tan \theta {)}^2\Rightarrow (a^2-b^2){\tan }^2\theta -2ac\tan \theta +c^2-b^2=0\text{It is given that}\alpha \text{~}and\beta \text{are the solutions of eq}(i)\text{Then,}\tan \alpha +\tan \beta =\frac{2ac}{a^2-b^2}\text{and}\tan \alpha \tan \beta =\frac{c^2-b^2}{a^2-b^2}\text{and}\tan \alpha \tan \beta =\frac{c^2-b^2}{a^2-b^2}\text{Now, LHS}=\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\frac{\frac{2ac}{a^2-b^2}}{1-\frac{c^2-b^2}{a^2-b^2}}=\frac{2ac}{a^2-b^2-c^2+b^2}=\frac{2ac}{a^2-c^2}$