Question

Prove that : $\frac{tan\Theta }{1-cot\Theta }+\frac{cot\Theta }{1-tan\Theta }=1+tan\Theta +cot\Theta$

Answer 1

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

solving LHS we get

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

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

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

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

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

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

$\Rightarrow 1+\tan \theta +\cot \theta =$RHS.