Answer
See the explanation below.
Work Step by Step
$\frac{\tan t}{\sec t-1}$
We multiply $\frac{\sec t+1}{\sec t+1}$ with the numerator and denominator of the provided expression:
$\begin{align}
& \frac{\tan t}{\sec t-1}=\frac{\tan t}{\sec t-1}.\frac{\sec t+1}{\sec t+1} \\
& =\frac{\tan t\left( \sec t+1 \right)}{{{\sec }^{2}}t-1}
\end{align}$
Applying the Pythagorean identity of trigonometry ${{\tan }^{2}}t={{\sec }^{2}}t-1$ , which comes out by solving $1+{{\tan }^{2}}t={{\sec }^{2}}t$ , then the above expression can be further simplified as:
$\begin{align}
& \frac{\tan t\left( \sec t+1 \right)}{{{\sec }^{2}}t-1}=\frac{\tan t\left( \sec t+1 \right)}{{{\tan }^{2}}t} \\
& =\frac{\sec t+1}{\tan t}
\end{align}$
Thus, the right side of the expression is equal to the left side, which is
$\frac{\sec t+1}{\tan t}=\frac{\tan t}{\sec t-1}$.
