Question

If $\alpha ,\beta$ are two values of $\theta$ satisfying the equation ${\sec }^2\theta +p\tan \theta +q=1$ then

• A. $tan(\alpha +\beta )=\frac{p}{q-1}$
• B. $tan(\alpha -\beta )=\frac{p}{q-1}$
• C. $tan(\alpha +\beta )=\frac{q}{p-1}$
• D. $tan(\alpha -\beta )=\frac{q}{p-1}$

Answer 1

The correct option is A $tan(\alpha +\beta )=\frac{p}{q-1}$

${\sec }^2\theta +p\tan \theta +q=1$

$\Rightarrow 1+{\tan }^2\theta +p\tan \theta +q=1\Rightarrow {\tan }^2\theta +p\tan \theta +q=0$

If $\alpha ,\beta$ are the roots of the given equation.then

$\tan \alpha +\tan \beta =-p$ and $\tan \alpha .\tan \beta =q$

$\therefore \tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha .\tan \beta }=\frac{p}{q-1}$