Answer
$$\frac{1}{2}$$
$$
y= \frac{1}{2}t+\frac{3}{2}
$$
Work Step by Step
Since
$f(t)=\dfrac{2}{1-t} .$ Then
\begin{align*}
f^{\prime}(-1)&=\lim _{h \rightarrow 0} \frac{f(-1+h)-f(-1)}{h}\\
&=\lim _{h \rightarrow 0} \frac{\frac{2}{1+1-h}-1}{h}\\
&=\lim _{h \rightarrow 0} \frac{\frac{2}{2+h}-1}{h}\\
&=\lim _{h \rightarrow 0} \frac{1}{2-h}\\
&=\frac{1}{2}\end{align*}
The tangent line at $a=-1$ is
$$
y=f^{\prime}(-1)(t+1)+f(-1)=\frac{1}{2}(t+1)+1=\frac{1}{2}t+\frac{3}{2}
$$
