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