Question

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

Answer 1

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

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

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

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

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

(proved)