Question

State whether the following statement is true or false.

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

• A. True
• B. False

Answer 1

The correct option is A True

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

$L.H.S=\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }$

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

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

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

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

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

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

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

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

$=R.H.S$

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