#### Answer

$\dfrac {\cot \alpha +1}{\cot \alpha -1}=\dfrac {1+\tan \alpha }{1-\tan \alpha }$

#### Work Step by Step

$\dfrac {\cot \alpha +1}{\cot \alpha -1}=\dfrac {\dfrac {\cos \alpha }{\sin \alpha }+1}{\dfrac {\cos \alpha }{\sin \alpha }-1}=\dfrac {\cos \alpha +\sin \alpha }{\cos \alpha -\sin \alpha }=\dfrac {\dfrac {\cos \alpha +\sin \alpha }{\cos \alpha }}{\dfrac {\cos \alpha -\sin \alpha }{\cos a}}=\dfrac {1+\tan \alpha }{1-\tan \alpha }$