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From the identity $\tan(x + y) = \dfrac{\tan(x) + \tan(y)}{1 - \tan(x)\tan(y)}$ we see that

$\tan \dfrac{(2^{16})! \pi}{2^{16} + 1} + \tan \dfrac{\pi}{2^{16} + 1} = \left(\tan \dfrac{((2^{16})! + 1)\pi}{2^{16} + 1}\right) \left(1 - \tan \dfrac{(2^{16})! \pi}{2^{16} + 1} \tan \dfrac{\pi}{2^{16} + 1}\right)$ , (A).

Now note that $2^{16} + 1$ is a Fermat prime . So by Wilson's Theorem , we have that

$((2^{16} + 1) - 1)! \equiv -1 \pmod{2^{16} + 1} \Longrightarrow (2^{16})! + 1 = n*(2^{16} + 1)$ for some integer $n$ .

Thus $\tan \dfrac{((2^{16})! + 1)\pi}{2^{16} + 1} = \tan (n*\pi) = 0$ , and therefore the given expression equals $\boxed{0}$ as well.

(Note that neither $\dfrac{(2^{16})!}{2^{16} + 1}$ nor $\dfrac{1}{2^{16} + 1}$ is an odd multiple of $\dfrac{1}{2}$ , so the second term in the product (A) is finite.)

Since $p = 2^{16} + 1$ is prime (Fermat), we know that $(p-1)! \equiv -1$ mod p (Wilson).

Thus we have $\frac{(2^{16})! \cdot \pi}{2^{16}+1} \equiv \frac{-\pi}{2^{16}+1} \ \ \text{mod}\ \pi.$ Because the tangent is periodic with period $\pi$ we have $\tan\left(\frac{(2^{16})! \cdot \pi}{2^{16}+1}\right) = \tan\left(\frac{-\pi}{2^{16}+1}\right),$ and from the oddness of the tangent function, $\tan (-x) = -\tan x$ , it follows that $\dots = -\tan\left(\frac{\pi}{2^{16}+1}\right).$ Since this is the additive opposite of the second term, the sum will be zero $\boxed{0}$ .