Answer
$$y =\frac{-3}{\pi}x+2$$
Work Step by Step
Given
$$x=\sin(\pi t),\ \ \ y=t^2+t \ \ \ ;(0,2)$$
Since at $(0,2)$, we have
\begin{aligned}x& =\sin(\pi t)\\
0&=\sin(\pi t )\\
\pi t&=0\\
t&=0,\ \pi ,\ 2\pi ,\cdots \end{aligned}
Then
$$ t=0,\ \ \ \ t=1,\ 2,\ 3,\cdots $$
Since $y=2$ at $t=1 $, then we reject all value execpt $t=1$
\begin{aligned}
\frac{d y}{d t}=2t+1, \ \ \ \ \frac{d x}{d t}=\pi \cos (\pi t)\end{aligned}
Then
\begin{aligned}
\frac{d y}{d x}&=\frac{d y / d t}{d x / d t}\\
&=\frac{2t+1}{\pi \cos (\pi t)}
\end{aligned}
Hence the slope at $(0,2)$
\begin{aligned}
m&= \frac{-3}{\pi}
\end{aligned}
The tangent line equation given by
\begin{aligned}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y-2}{x-0}&=\frac{-3}{\pi}\\
y-2&= \frac{-3}{\pi}x\\
y&=\frac{-3}{\pi}x+2
\end{aligned}
