## Trigonometry (11th Edition) Clone

Express $\tan\Big(\frac{3\pi}{4}+x\Big)$ as a function of $x$ alone. $$\tan\Big(\frac{3\pi}{4}+x\Big)=\frac{\tan x-1}{1+\tan x}$$
$$\tan\Big(\frac{3\pi}{4}+x\Big)$$ 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.