## Trigonometry (11th Edition) Clone

$$\tan3x=\frac{3\tan x-\tan^3x}{1-3\tan^2x}$$
$$\tan3x$$ The question asks to write $\tan3x$ in terms of a trigonometric function of $x$. $$\tan3x=\tan(2x+x)$$ - Sum Identity for tangent: $\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$ with here $A=2x$ and $B=x$ $$\tan3x=\frac{\tan2x+\tan x}{1-\tan2x\tan x}$$ - For $\tan2x$, we apply Double-Angle Identity for tangent: $\tan2A=\frac{2\tan A}{1-\tan^2A}$ with $A=x$. $$\tan3x=\frac{\frac{2\tan x}{1-\tan^2x}+\tan x}{1-\frac{2\tan x}{1-\tan^2x}\times\tan x}$$ $$\tan3x=\frac{\frac{2\tan x+\tan x(1-\tan^2x)}{1-\tan^2x}}{\frac{1-\tan^2x-2\tan^2x}{1-\tan^2x}}$$ We can eliminate both $1-\tan^2x$ here. $$\tan3x=\frac{2\tan x+\tan x(1-\tan^2x)}{1-\tan^2x-2\tan^2x}$$ $$\tan3x=\frac{2\tan x+\tan x-\tan^3x}{1-3\tan^2x}$$ $$\tan3x=\frac{3\tan x-\tan^3x}{1-3\tan^2x}$$ That is the result we need to find.