Question

$\frac{tan\Theta }{1-cot\Theta }+\frac{cot\Theta }{1-tan\Theta }=1+sec\Theta cosec\Theta$.

Answer 1

Consider the problem

$\begin{array}{c}\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }\\ =\frac{\tan \theta }{1-\frac{1}{\tan \theta }}+\frac{\frac{1}{\tan \theta }}{1-\tan \theta }\\ =\frac{\tan \theta }{\frac{\tan \theta -1}{\tan \theta }}+\frac{1}{\tan \theta \left(1-\tan \theta \right)}\\ =\frac{{\tan }^2\theta }{\tan \theta -1}-\frac{1}{\tan \theta \left(\tan \theta -1\right)}\\ =\frac{{\tan }^3\theta -1}{\tan \theta \left(\tan \theta -1\right)}\\ =\frac{\left(\tan \theta -1\right)\left({\tan }^2\theta +\tan \theta +1\right)}{\tan \theta \left(\tan \theta -1\right)}\left[\because a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\right]\\ =\frac{{\tan }^2\theta +\tan \theta +1}{\tan \theta }\\ =\tan \theta +1+\cot \theta \\ =\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }+1\\ =\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }+1\\ =\frac{1}{\sin \theta \cos \theta }+1\\ =\sec \theta \cos ec\theta +1\end{array}$

$LHS=RHS$