Question

Simplify:-

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

Answer 1

We have,

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

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

Since,

${(a+b)}^2=a^2+b^2+2ab$

${(a-b)}^2=a^2+b^2-2ab$

$a^2-b^2=(a+b)(a-b)$

Therefore,

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

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

Since,

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

Therefore,

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

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

Hence, proved.