Question

Prove that following identities:
(i) $\frac{\sin \theta -\cos \theta }{sin\theta +\cos \theta }+\frac{\sin \theta +\cos \theta }{\sin \theta -\cos \theta }=\frac{2}{2si{n}^2\theta -1}$
(ii)$\frac{\sin \theta +\cos \theta }{\sin \theta -\cos \theta }+\frac{\sin \theta -\cos \theta }{\sin \theta +\cos \theta }=\frac{2}{1-2{\cos }^2\theta }$

Answer 1

(i) Consider LHS$=\frac{\sin \theta -\cos \theta }{\sin \theta +cos\theta }+\frac{sin\theta +cos\theta }{sin\theta -\cos \theta }$
$=\frac{(\sin \theta -\cos \theta {)}^2+(\sin \theta +\cos \theta {)}^2}{(\sin \theta +\cos \theta )(\sin \theta -\cos \theta )}$
$=\frac{{\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta +{\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta }{{\sin }^2\theta -co{s}^2\theta }$
$=\frac{1+1}{{\sin }^2\theta -(1-{\sin }^2\theta )}$[Since, ${\sin }^2\theta +{\cos }^2\theta =1$]
$=\frac{2}{{\sin }^2\theta -1+{\sin }^2\theta }$
$=\frac{2}{2si{n}^2\theta -1}$
$=$RHS
Hence,$\frac{\sin \theta -\cos \theta }{sin\theta +\cos \theta }+\frac{\sin \theta +\cos \theta }{\sin \theta -\cos \theta }=\frac{2}{2si{n}^2\theta -1}$
(ii) LHS$=\frac{\sin \theta +\cos \theta }{\sin \theta -\cos \theta }+\frac{\sin \theta -\cos \theta }{\sin \theta +\cos \theta }$
$=\frac{(\sin \theta +\cos \theta {)}^2+(\sin \theta -\cos \theta {)}^2}{(\sin \theta -\cos \theta )(\sin \theta +\cos \theta )}$
$=\frac{{\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta +{\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta }{(si{n}^2\theta -{\cos }^2\theta )}$
$=\frac{1+1}{(1-co{s}^2\theta )-co{s}^2\theta }$
(since, ${\sin }^2\theta +{\cos }^2\theta =1$)
$=\frac{2}{1-2{\cos }^2\theta }$
$=$RHS
Hence, $\frac{\sin \theta +\cos \theta }{\sin \theta -\cos \theta }+\frac{\sin \theta -\cos \theta }{\sin \theta +\cos \theta }=\frac{2}{1-2{\cos }^2\theta }$