Question

Prove that $\frac{\sin \theta +\cos \theta }{\sin \theta -\cos \theta }+\frac{\sin \theta -\cos \theta }{\sin \theta +\cos \theta }=\frac{2}{2{\sin }^2\theta -1}=\frac{2}{1-2{\cos }^2\theta }$

Answer 1

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

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

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

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

$=\frac{2{\sin }^2\theta +2{\cos }^2\theta }{{\sin }^2\theta -(1-{\sin }^2\theta )}$

$=\frac{2({\sin }^2\theta +{\cos }^2\theta )}{2{\sin }^2\theta -1}$

$=\frac{2}{2{\sin }^2\theta -1}$ ........$\left(1\right)$

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

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

$=\frac{2}{1-2{\cos }^2\theta }$ ........$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get

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

Hence proved.