Question

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

Answer 1

$L.H.S=$ $\frac{1+\sec \theta -\tan \theta }{1+\sec \theta +\tan \theta }$

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

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

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

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

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

$=\frac{1-\sin \theta }{\cos \theta }$ $=R.H.S$

Hence proved.