Question

Prove that:

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

Answer 1

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

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

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

and

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

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

$=1+\sec \theta \csc \theta$

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