Answer
$$\frac{\tan\frac{5\pi}{12}+\tan\frac{\pi}{4}}{1-\tan\frac{5\pi}{12}\tan\frac{\pi}{4}}=-\sqrt3$$
Work Step by Step
$$X=\frac{\tan\frac{5\pi}{12}+\tan\frac{\pi}{4}}{1-\tan\frac{5\pi}{12}\tan\frac{\pi}{4}}$$
From the identity of the sum of tangent:
$$\frac{\tan A+\tan B}{1-\tan A\tan B}=\tan(A+B)$$
So here $X$ actually follows the above identity with $A=\frac{5\pi}{12}$ and $B=\frac{\pi}{4}$.
Therefore, $X$ can also be rewritten as
$$X=\tan\Big(\frac{5\pi}{12}+\frac{\pi}{4}\Big)$$
$$X=\tan\frac{5\pi+3\pi}{12}$$
$$X=\tan\frac{8\pi}{12}=\tan\frac{2\pi}{3}$$
From Section 5.3, as cofunction identities:
$$\tan\theta=\cot\Big(\frac{\pi}{2}-\theta\Big)$$
Therefore, $$X=\cot\Big(\frac{\pi}{2}-\frac{2\pi}{3}\Big)$$
$$X=\cot\Big(-\frac{\pi}{6}\Big)$$
Also, we know $$\cot(-\theta)=-\cot\theta$$
$$X=-\cot\Big(\frac{\pi}{6}\Big)$$
$$X=-\sqrt3$$
which means
$$\frac{\tan\frac{5\pi}{12}+\tan\frac{\pi}{4}}{1-\tan\frac{5\pi}{12}\tan\frac{\pi}{4}}=-\sqrt3$$
