Question

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

Answer 1

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

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

$=\frac{{\tan }^2\theta }{\tan \theta -1}-\frac{\frac{1}{\tan \theta }}{\tan \theta -1}$

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

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

$=\frac{{\tan }^2\theta +\tan \theta +1}{\tan \theta }$

$=\tan \theta +1+\cot \theta$

$=\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }+1$

$=1+\frac{{\sin }^2\theta +{\cos }^2\theta }{\cos \theta \cdot \sin \theta }$

$=1+\frac{1}{\cos \theta \sin \theta }$

$=1+\sec \theta \cdot cosec\theta$.