Question

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

Answer 1

L.H.S

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

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

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

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

We know that

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

Therefore,

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

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

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

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

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

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

Hence, proved.