Question

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

Answer 1

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

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

$=\frac{{\sin }^2\theta }{\cos \theta (\sin \theta -\cos \theta )}+\frac{{\cos }^2\theta }{\sin \theta (\cos \theta -\sin \theta )}$

$=\frac{{\sin }^3\theta -{\cos }^3\theta }{\sin \theta \cos \theta (\sin \theta -\cos \theta )}$

$=\frac{(\sin \theta -\cos \theta )({\sin }^2\theta +\sin \theta \cos \theta +{\cos }^2\theta )}{\sin \theta \cos \theta (\sin \theta -\cos \theta )}$

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

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

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

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

Hence proved.