Question

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

Answer 1

L.H.S. $=\frac{1+\sin 2\theta }{1-\sin 2\theta }$
$=\frac{{\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta }{{\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta }$
$={\left(\frac{\sin \theta +\cos \theta }{\sin \theta -\cos \theta }\right)}^2={\left(\frac{1+\tan \theta }{1-\tan \theta }\right)}^2$ (Dividing numerator and denominator by $\cos \theta$)
Ans: 2