Question

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

Answer 1

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

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

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

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

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

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

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

$(\because a^3-b^3=(a-b\left)\right(a^2+ab+b^2\left)\right)$

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

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

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