Is the identity $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\tan \theta +\cot \theta$ true or false?

The correct option is A (True)

LHS :

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

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

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

$=\frac{-{\tan }^2\theta +\frac{1}{\tan \theta }}{1-\tan \theta }$
$=\frac{\frac{-{\tan }^3\theta +1}{\tan \theta }}{1-\tan \theta }$

$=-\frac{(1-{\tan }^3\theta )}{\tan \theta (1-\tan \theta )}$

$=\frac{(1-\tan \theta )(1+\tan \theta +{\tan }^2\theta )}{\tan \theta (1-\tan \theta )}$
(Since, $a^3-b^3=(a-b)(a^2+ab+b^2$))

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

$=1+\cot \theta +\tan \theta$ = RHS
Hence, the given identity is true.