Question

The angle made by the latus-rectum of the parabola $y^2=4ax$ at it's vertex is $\theta$ then $3|\tan (\frac{\pi }{4}+\frac{\theta }{2})+\tan (\frac{\pi }{4}-\frac{\theta }{2})|$ is

Answer 1

$\tan \frac{\theta }{2}=2\Rightarrow \tan (\frac{\pi }{4}+\frac{\theta }{2})=\frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}}=-3$
$\tan (\frac{\pi }{4}-\frac{\theta }{2})=\frac{1-\tan \frac{\theta }{2}}{1+\tan \frac{\theta }{2}}=-\frac{1}{3}$
$\therefore 3|\tan (\frac{\pi }{4}+\frac{\theta }{2})+\tan (\frac{\pi }{4}-\frac{\theta }{2})|=10$