#### Answer

$$L(t)=2\pi (t-1/4) -1 $$

#### Work Step by Step

Given $$R(t)=\tan \left(\pi\left(t-\frac{1}{2}\right)\right), \quad a=\frac{1}{4}$$
Since
\begin{align*}
R'(t)&=\pi \sec^2 \left(\pi\left(t-\frac{1}{2}\right)\right)\\
R(1/4)&= 2\pi
\end{align*}
Then the linear approximation is given by
\begin{align*}
L(t)&=R^{\prime}(a)(t-a)+R(a)\\
&=2\pi (t-1/4) -1
\end{align*}