Question

Prove that: $\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}=\frac{1+\sin \theta }{\cos \theta }$.

Answer 1

Consider the following question.

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

Consider, $LHS=\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}$

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

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

$=(\tan \theta +\sec \theta )$

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

Hence, this is the required answer.