Answer
See the explanation below.
Work Step by Step
The expression on the left side $\frac{\sin t-1}{\text{cos }t}$ can be further simplified by multiplying by cot t.
$\begin{align}
& \frac{\sin t-1}{\text{cos }t}=\frac{\sin t-1}{\text{cos }t}.\frac{\cot t}{\cot t} \\
& =\frac{\sin t.\cot t-\cot t}{\cos t.\cot t}
\end{align}$
We know the quotient identity of $\cot x=\frac{\cos x}{\sin x}$. Applying the identity to the expression, we get:
$\begin{align}
& \frac{\sin t\cot t-\cot t}{\cos t.\cot t}=\frac{\sin t\frac{\cos t}{\sin t}-\cot t}{\cos t.\cot t} \\
& =\frac{1.\cos t-\cot t}{\cos t.\cot t} \\
& =\frac{\cos t-\cot t}{\cos t.\cot t}
\end{align}$
Hence, the left side is equal to the right side $\frac{\sin t-1}{\text{cos }t}=\frac{\cos t-\cot t}{\cos t\cot t}$.
