Answer
$$\frac{1}{1+\sin t}+\frac{1}{1-\sin t} =2\sec^2 t$$
Work Step by Step
Since
\begin{align*}
\frac{1}{1+\sin t}+\frac{1}{1-\sin t}&=\frac{1-\sin t+1+\sin t}{(1+\sin t)(1-\sin t)}\\
&=\frac{2 }{ 1-\sin^2 t },\ \ \text{use}\ \ \cos ^2t+\sin ^2t=1 \\
&=\frac{2 }{ \cos^2t }\\
&=2\sec^2 t
\end{align*}
Then
$$\frac{1}{1+\sin t}+\frac{1}{1-\sin t} =2\sec^2 t$$
