The digits needed are actually the 101st digit and 102nd digit after the decimal point of $1\over\pi$. These two digits happen to be "12" (courtesy of Wolframalpha) so $\tan\left(10^{100}\right)\approx\tan(0.12\pi)$.

To see why this works, let's follow the entire calculation:

Fetching the first 102 digits of $1\over\pi$ from Wolframalpha, we get:

0.318309886183790671537767526745028724068919291480912897495334688117793595268453070180227605532506171912... (note the "12" in the end)

So $10^{100}\over\pi$ is:

3183098861837906715377675267450287240689192914809128974953346881177935952684530701802276055325061719.12...

Which we will write as $N+0.12...$ (where $N$ is an integer).

Now we can write:

$\tan\left(10^{100}\right)=\tan\left[\left(10^{100}\over\pi\right)\cdot\pi\right]\approx\tan\left[\left(N+0.12\right)\cdot\pi\right]=\tan\left(N\pi+0.12\pi\right)$

And since $\tan(x)$ has a period of $\pi$, we get:

$\tan(N\pi+0.12\pi)=\tan(0.12\pi)$

Which can further be approximated as:

$\tan(0.12\pi)\approx\tan(0.38)\approx0.38$
(the last step uses the approximation of $\tan(x)\approx{x}$ for small $x$)