Question

Solve:
$\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\tan \theta +\cot \theta$

Answer 1

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

LHS

$\frac{{\tan }^2\theta }{(\tan \theta -1)}-\frac{\cot \theta }{(\tan \theta -1)}=\frac{{\tan }^3\theta -1}{\tan \theta (\tan \theta -1)}=\frac{{\tan }^3\theta -1}{{\tan }^2\theta -\tan \theta }$

$[a^3-b^3-(a-b\left)\right(a^2+ab+b^2]$

$\Rightarrow \frac{(\tan \theta -1)({\tan }^3\theta -1+\tan \theta )}{\left(\tan \theta \right)(\tan \theta -1)}=1+\frac{1}{\tan \theta }+\tan \theta =1+\tan \theta +\cot \theta$=RHS