Chapter 7 - 7.5 - Multiple-Angle and Product-to-Sum Formulas - 7.5 Exercises - Page 548: 26

$tan~3x=\frac{-tan^3x+3~tan~x}{-3~tan^2x+1}=\frac{tan^3x-3~tan~x}{3~tan^2x-1}$

$tan~3x=tan~(x+2x)=\frac{tan~x+tan~2x}{1-tan~x~tan~2x}=\frac{tan~x+\frac{2~tan~x}{1-tan^2x}}{1-tan~x~\frac{2~tan~x}{1-tan^2x}}=\frac{\frac{tan~x-tan^3x+2~tan~x}{1-tan^2x}}{\frac{1-tan^2x-2~tan^2x}{1-tan^2x}}=\frac{-tan^3x+3~tan~x}{-3~tan^2x+1}=\frac{tan^3x-3~tan~x}{3~tan^2x-1}$