Question

plzz solve it ......​

Answer 1

Step-by-step explanation:

$\frac{ \tan( x) }{1 - \cot(x) } + \frac{ \cot(x) }{1 - \tan(x) } = > \frac{ \tan(x) }{1 - \frac{1}{ \tan(x) }} +$

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

$= > \frac{ \frac{1}{ \tan(x) } - {( \tan(x)) }^{2} }{ 1 - \tan(x)} = > \frac{1 - {( \tan(x)) }^{3} }{(1 - \tan(x)) \tan(x) }$

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

$\frac{1 + {( \tan(x)) }^{2} + \tan(x) }{ \tan(x)} = > 1 + \frac{1 + {( \tan(x)) }^{2} }{ \tan(x)} = >$

$= > 1 + \frac{ {( \sec(x)) }^{2} }{ \tan(x)} = > 1 + \sec(x). \frac{ \sec(x) }{ \tan(x)}$

$= > 1 + \sec(x) \csc(x)$

Hence proved. Hope it helps you.

Answer 2

This is ur answer....
Pls mark as brainliest...