# Question #a4b0f

##### 1 Answer
Oct 8, 2017

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

#### Explanation:

We have to prove, $\tan \frac{\theta}{1 - \cot \theta} + \cot \frac{\theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$

Let us take Left Hand Side (L.H.S.)
$\Rightarrow \tan \frac{\theta}{1 - \cot \theta} + \cot \frac{\theta}{1 - \tan \theta}$

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

$\Rightarrow \frac{\sin \frac{\theta}{\cos} \theta}{\frac{\sin \theta - \cos \theta}{\sin} \theta} + \frac{\cos \frac{\theta}{\sin} \theta}{\frac{\cos \theta - \sin \theta}{\cos} \theta}$

$\Rightarrow \frac{\sin \frac{\theta}{\cos} \theta . \sin \theta}{\sin \theta - \cos \theta} + \frac{\cos \frac{\theta}{\sin} \theta . \cos \theta}{- \left(\sin \theta - \cos \theta\right)}$

$\Rightarrow \frac{{\sin}^{2} \frac{\theta}{\cos} \theta}{\sin \theta - \cos \theta} - \frac{{\cos}^{2} \frac{\theta}{\sin} \theta}{\sin \theta - \cos \theta}$

$\Rightarrow \frac{{\sin}^{2} \frac{\theta}{\cos} \theta - {\cos}^{2} \frac{\theta}{\sin} \theta}{\sin \theta - \cos \theta}$

$\Rightarrow \frac{\frac{{\sin}^{3} \theta - {\cos}^{3} \theta}{\sin \theta \cos \theta}}{\sin \theta - \cos \theta}$

$\Rightarrow \frac{\left(\sin \theta - \cos \theta\right) \left({\sin}^{2} \theta + \sin \theta . \cos \theta + {\cos}^{2} \theta\right)}{\sin \theta \cos \theta} . \frac{1}{\sin \theta - \cos \theta}$

$\Rightarrow {\sin}^{2} \frac{\theta}{\sin \theta \cos \theta} + \frac{\sin \theta . \cos \theta}{\sin \theta . \cos \theta} + {\cos}^{2} \frac{\theta}{\sin \theta . \cos \theta}$

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

$\Rightarrow \tan \theta + 1 + \cot \theta$ = L. H. S.