Answer
Express $\tan\Big(\frac{3\pi}{4}+x\Big)$ as a function of $x$ alone.
$$\tan\Big(\frac{3\pi}{4}+x\Big)=\frac{\tan x-1}{1+\tan x}$$
Work Step by Step
$$\tan\Big(\frac{3\pi}{4}+x\Big)$$
To be expressed as a function of $x$ alone, we need here the identity for sum of tangents:
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
Apply it to $\tan\Big(\frac{3\pi}{4}+x\Big)$, we have
$$\tan\Big(\frac{3\pi}{4}+x\Big)=\frac{\tan\frac{3\pi}{4}+\tan x}{1-\tan\frac{3\pi}{4}\tan x}$$
Now we need to calculate $\tan\frac{3\pi}{4}$.
$$\tan\Big(\frac{3\pi}{4}\Big)=\tan\Big(\pi-\frac{\pi}{4}\Big)=\frac{\tan\pi-\tan\frac{\pi}{4}}{1+\tan\pi\tan\frac{\pi}{4}}$$
$$\tan\frac{3\pi}{4}=\frac{0-1}{1-0\times1}=\frac{-1}{1}=-1$$
Apply the result back to $\tan\Big(\frac{3\pi}{4}+x\Big)$
$$\tan\Big(\frac{3\pi}{4}+x\Big)=\frac{-1+\tan x}{1-(-1)\times\tan x}$$
$$\tan\Big(\frac{3\pi}{4}+x\Big)=\frac{\tan x-1}{1+\tan x}$$
This is the ultimate function we need to find.
