Question

Find
$\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=cosec\theta \sec \theta +1$

Answer 1

$\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=\sec \theta \cdot cosec\theta +1$

L.H.S$=\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }$

$=\frac{\tan \theta }{1-\frac{1}{\tan \theta }}+\frac{\frac{1}{\tan \theta }}{1-\tan \theta }$

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

$=\frac{1}{-1+\tan \theta }[{\tan }^2\theta -\frac{1}{\tan \theta }]$

$=\frac{1}{\tan \theta -1}\cdot \frac{{\tan }^3\theta -1}{\tan \theta }$ $[\because (a^3-b^3)=(a-b\left)\right(a^2+b^2+ab\left)\right]$

$=\frac{1}{\tan \theta -1}\cdot (\tan \theta -1)\frac{{\tan }^2\theta +1+\tan \theta }{\tan \theta }$

$=\frac{{\sec }^2\theta +\tan \theta }{\tan \theta }$

$=\frac{{\sec }^2\theta }{\tan \theta }+\frac{\tan \theta }{\tan \theta }$

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

$=\frac{1}{\sin \theta \cdot \cos \theta }+1$

$=sec\theta \cdot cosec\theta +1$

$=$ R.H.S

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