Answer
$\dfrac {1}{sec\alpha -\tan \alpha }=sec\alpha +\tan \alpha $
Work Step by Step
$\dfrac {1}{sec\alpha -\tan \alpha }=\dfrac {1}{\dfrac {1}{\cos \alpha }-\dfrac {\sin \alpha }{\cos \alpha }}=\dfrac {\cos \alpha }{1-\sin \alpha }=\dfrac {\cos \alpha \left( 1+\sin \alpha \right) }{\left( 1-\sin \alpha \right) \left( 1+\sin \alpha \right) }=\dfrac {\cos \alpha \left( 1+\sin \alpha \right) }{1-\sin ^{2}\alpha }=\dfrac {\cos \alpha \left( 1+\sin \alpha \right) }{\cos ^{2}\alpha }=\dfrac {\left( 1+\sin \alpha \right) }{\cos \alpha }=sec\alpha +\tan \alpha $
