#### Answer

$L(x)=x-\pi$

#### Work Step by Step

$f(x)=\tan{x} $,
$L(x)=f(a)+f'(a)(x-a)$
$f'(x)=sec^2{x}$
$a=\pi$
$f(\pi)=\tan{\pi}=0$
$f'(\pi)=\sec^2{\pi}=(-1)^2=1$
then $L(x)=0+(x-\pi)=x-\pi$
$L(x)=x-\pi$
Thus, the final answer is: $L(x)=x-\pi$