Question

Prove that $\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }$, using the identify.

Answer 1

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

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

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

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

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

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

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

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

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