Question

Prove that:

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

Answer 1

STEP 1 : Solving the Left Hand Side (LHS) of the equation

Taking the LHS and solving we get,

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

$=\frac{\tan \theta }{\left(1-{\displaystyle \frac{1}{\tan \theta }}\right)}+\frac{{\displaystyle \frac{1}{\tan \theta }}}{(1-\tan \theta )}$

$=\frac{\tan \theta }{{\displaystyle \frac{\left(\tan \theta -{\displaystyle 1}\right)}{\tan \theta }}}+\frac{{\displaystyle 1}}{\tan \theta (1-\tan \theta )}$

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

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

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

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

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

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

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

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

i.e. $\frac{\tan \theta }{(1-cot\theta )}+\frac{cot\theta }{(1-\tan \theta )}=1+\tan \theta +cot\theta$

Hence proved.