Question

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

Answer 1

To prove:

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

Solution:

L.H.S. $=\frac{\sec \theta -\tan \theta }{\sec \theta +\tan \theta }$

Multiplying numerator and denominator by $\sec \theta -tan\theta ,$

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

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

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

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

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

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

$=$ R.H.S.

Hence proved.