Question

Solve
$\frac{\sin \theta +\cos \theta }{\sin \theta -\cos \theta }+\frac{\sin \theta -\cos \theta }{\sin \theta +\cos \theta }=\frac{2{\sec }^2\theta }{{\tan }^2\theta -1}$

Answer 1

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

$L.H.S$

$=\frac{{(\sin \theta +\cos \theta )}^2+{(\sin \theta -\cos \theta )}^2}{(\sin \theta +\cos \theta )(\sin \theta -\cos \theta )}$

$=\frac{{\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta +{\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta }{{\sin }^2\theta -{\cos }^2\theta }$

$=\frac{{\sin }^2\theta +{\cos }^2\theta +{\sin }^2\theta +{\cos }^2\theta }{{\sin }^2\theta -{\cos }^2\theta }$

$=\frac{2{\sin }^2\theta +2{\cos }^2\theta }{{\sin }^2\theta -{\cos }^2\theta }$

$=\frac{2({\sin }^2\theta +{\cos }^2\theta )}{({\sin }^2\theta -{\cos }^2\theta )}$

Now, $RHS$

$=\frac{2{\sec }^2\theta }{{\tan }^2\theta -1}=\frac{2\frac{1}{{\cos }^2\theta }}{\frac{{\sin }^2\theta }{{\cos }^2\theta }-1}$

$=\frac{2/{\cos }^2\theta }{{\sin }^2\theta -{\cos }^2\theta /{\cos }^2\theta }$

$=\frac{2}{{\sin }^2\theta -{\cos }^2\theta }$

$\therefore L.H.S=R.H.S$