Question

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

Answer 1

Given,

$L.H.S=\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=\frac{\tan \theta }{1-\frac{1}{\tan \theta }}+\frac{\cot \theta }{1-\tan \theta }=\frac{\tan \theta }{\frac{\tan \theta -1}{\tan \theta }}+\frac{\cot \theta }{1-\tan \theta }=\frac{{\tan }^2\theta }{\tan \theta -1}+\frac{\cot \theta }{1-\tan \theta }=\frac{-{\tan }^2\theta }{1-\tan \theta }+\frac{\cot \theta }{1-\tan \theta }=\frac{-{\tan }^2\theta +\cot \theta }{1-\tan \theta }=\frac{-{\tan }^2\theta +\frac{1}{\tan \theta }}{1-\tan \theta }=\frac{\frac{-{\tan }^3\theta +1}{\tan \theta }}{1-\tan \theta }=\frac{1-{\tan }^3\theta }{\tan \theta (1-\tan \theta )}=\frac{(1-\tan \theta )(1+\tan \theta +{\tan }^2\theta )}{\tan \theta (1-\tan \theta )}=\frac{1+\tan \theta +{\tan }^2\theta }{\tan \theta }=\frac{1}{\tan \theta }+\frac{\tan \theta }{\tan \theta }+\frac{{\tan }^2\theta }{\tan \theta }=\cot \theta +1+\tan \theta =1+\tan \theta +\cot \theta =R.H.S$