Question

Proove:

$\sqrt{\frac{1-sin\Theta }{1+sin\Theta }}=sec\Theta -tan\Theta$

Answer 1

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

L.H.S: $\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}$

Dividing by $\cos \theta$ in

numerator and denominator

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

Rationalising, we get

$\sqrt{\frac{\sec \theta -\tan \theta \times \sec \theta -\tan \theta }{(\sec \theta +\tan \theta )\times \sec \theta -\tan \theta }}$

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

$=\sec \theta -\tan \theta =$R.H.S.