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