Question

Prove that $\frac{tan\theta }{1-cot\theta }+\frac{cot\theta }{1-tan\theta }=1+tan\theta +cot\theta =sec\theta cosec\theta +1$.

Answer 1

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

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

$\Rightarrow \frac{(\tan \theta +\cot \theta +1)(2-(\tan \theta +\cot \theta \left)\right)}{2-(\tan \theta +\cot \theta )}=1+\tan \theta +\cot \theta$

$\Rightarrow 1+\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }=1+\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }=1+\frac{1}{\sin \theta \cos \theta }=1+\sec \theta \csc \theta$