Answer
$(\csc{\theta}+\cot{\theta})(\sec{\theta}-1)=\tan{\theta}$
Work Step by Step
Use agraphing utility to graph $y=(\csc{\theta}+\cot{\theta})(\sec{\theta}-1)$.
Refer to the graph below.
Notice that the graph looks the same as that of $y=\tan{\theta}$.
RECALL:
(1) $\csc{\theta}=\dfrac{1}{\sin{\theta}}\\\\$
(2) $\sec{\theta}=\dfrac{1}{\sin{\theta}}\\\\$
(3) $\cot{\theta}=\dfrac{\cos{\theta}}{\sin{\theta}}\\\\$
(4) $\tan{\theta}=\dfrac{\sin{\theta}}{\cos{\theta}}\\\\$
Use the definitions above to obtain:
\begin{align*}
(\csc{\theta}+\cot{\theta})(\sec{\theta}-1)&=\left(\frac{1}{\sin{\theta}}+\frac{\cos{\theta}}{\sin{\theta}}\right)\left(\frac{1}{\cos{\theta}}-1\right)\\\\
&=\left(\frac{1+\cos{\theta}}{\sin{\theta}}\right)\left(\frac{1}{\cos{\theta}}-\frac{\cos{\theta}}{\cos{\theta}}\right)\\\\
&=\left(\frac{1+\cos{\theta}}{\sin{\theta}}\right)\left(\frac{1-\cos{\theta}}{\cos{\theta}}\right)\\\\
&=\left(\frac{1-\cos^2{\theta}}{\sin{\theta}\cos{\theta}}\right)\\\\
&=\frac{\sin^2{\theta}}{\sin{\theta}\cos{\theta}}\\\\
&=\frac{\sin{\theta}}{\cos{\theta}}\\\\
&=\tan{\theta}
\end{align*}
Therefore,
$$(\csc{\theta}+\cot{\theta})(\sec{\theta}-1)=\tan{\theta}$$
