Question

Prove the following:
$\frac{(\tan \theta +\sec \theta -1)}{\tan \theta -\sec \theta +1)}=\frac{1+\sin \theta }{\cos \theta }$.

Answer 1

LHS

$\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}$

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

$\frac{\tan \theta +\sec \theta -(\sec \theta -\tan \theta )(\sec \theta +\tan \theta )}{\tan \theta -\sec \theta +1}$

$\frac{\tan \theta +\sec \theta (1-\sec \theta +\tan \theta )}{\tan \theta -\sec \theta +1}$

$\tan \theta +\sec \theta \frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }\frac{1+\sin \theta }{\cos \theta }$