Answer
a) The value is $0$.
b) The value is $0$.
Work Step by Step
(a)
As $\tan \left( t \right)=\frac{\sin \left( t \right)}{\cos \left( t \right)}$
So,
$\tan \left( \pi \right)=\frac{\sin \left( \pi \right)}{\cos \left( \pi \right)}$ ,
In the unit circle, the point corresponding to $t=\frac{\pi }{2}$ has the coordinates $\left( -1,0 \right)$. Therefore, use $x=-1$ and $y=0$, such that,
$\sin \left( \pi \right)=y=0\text{ }$ and $\cos \left( \pi \right)=x=-1$
So,
$\begin{align}
& \tan \left( \pi \right)=\frac{\sin \left( \pi \right)}{\cos \left( \pi \right)} \\
& =\frac{0}{-1} \\
& =0
\end{align}$
Thus, the value of the trigonometric function $\tan \left( \pi \right)$ is $0$.
(b)
We know that the periodic properties of sine and cosine functions are,
$\tan \left( t+\pi \right)=\tan \left( t \right)$ and $\text{cot}\left( t+\pi \right)=\cot \left( t \right)$.
Therefore,
$\begin{align}
& \tan \left( 17\pi \right)=\tan \left( \pi +16\pi \right) \\
& =\tan \left( \pi \right)
\end{align}$
Now from $\tan \left( \pi \right)=0$. So,
$\tan \left( 17\pi \right)=0$
Thus, the value of the trigonometric function $\tan \left( 17\pi \right)$ is $0$.
