Question

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

Answer 1

To prove:- $\frac{\sec \theta +\tan \theta -1}{\tan \theta -\sec \theta +1}=\tan \theta +\tan \theta$

Proof:-

Taking L.H.S.-

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

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

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

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

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

$=$ R.H.S.

Hence proved