Question

Prove that
$\tan (\frac{\pi }{4}+\frac{\theta }{2})=\frac{1}{\sec \theta -tan\theta }$

Answer 1

$\tan (\frac{\pi }{4}+\frac{\theta }{2})=\frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}}$ ………$\left(1\right)$

$\frac{1}{\sec \theta -\tan \theta }=\frac{\sec \theta +\tan \theta }{{\sec }^2\theta -{\tan }^2\theta }$

$=\sec \theta +\tan \theta$

$=\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }=\frac{1+\sin \theta }{\cos \theta }$

$=\frac{1+\frac{2\tan \frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}}{\frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}}$

$=\frac{(1+\tan \frac{\theta }{2}{)}^2}{1-{\tan }^2\frac{\theta }{2}}$

$=\frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}}$ ……….$\left(2\right)$

$\therefore$ From $\left(1\right)$ & $\left(2\right)$ $1+$ is proved.