Originally Posted by
Chris L T521 Recall that $\displaystyle L(x)= f(x_0)+f'(x_0)(x-x_0)$
$\displaystyle \because f(x)=\tan(x),~f'(x)=\sec^2(x)$
Thus, $\displaystyle f(\pi)=\tan(\pi)=0$ and $\displaystyle f'(\pi)=sec^2(\pi)=1$
Therefore, $\displaystyle L(x)=\dots$
Can you take it from here?