Question

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

Answer 1

We consider $\frac{\sec \theta -\tan \theta }{\sec \theta +\tan \theta }$

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

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

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

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

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

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