How would you find $\tan(3\cdot\arctan(x))$? How would you find $\tan(3\cdot\tan^{-1}(x))$ as quickly as possible?
I don't really understand how to use $\tan(2x)$ or any other identity for that matter to solve this.
 A: Begin with $$\tan(2.\arctan(x))=\frac{2\tan(\arctan(x))}{1-\tan^2(\arctan(x))}=\frac{2x}{1-x^2}.$$
Next, 
$$\begin{align}\tan(3.\arctan(x))&=\tan(2\arctan(x)+\arctan(x))\\&=\frac{\tan(2\arctan(x))+\tan(\arctan(x))}{1-\tan(2\arctan(x))\tan(\arctan(x))}\\&=\frac{\frac{2x}{1-x^2}+x}{1-\frac{2x}{1-x^2}x}\\&=\frac{3x-x^3}{1-3x^2}\end{align}$$
A more general solution is obtained combining the binomial and Euler's formulas:
$$(1+i\tan(\theta))^3=(1+ix)^3=1+3ix-3x^2-ix^3=(\frac{\cos\theta+i\sin\theta}{cos\theta})^3=\frac{\cos3\theta+i\sin3\theta}{\cos^3\theta},$$
so that, by equating the ratios of imaginary parts over real parts,
$$\tan(3\arctan x)=\tan(3\theta)=\frac{3x-x^3}{1-3x^2}.$$
Similarly, $(1+ix)^4=1+4ix-6x^2-4ix^3+x^4$ gives
$$\tan(4\arctan(x))=\frac{4x-4x^3}{1-6x^2+x^4},$$
$(1+ix)^5=1+5ix-10x^2-10ix^3+5x^4+ix^5,$
$$\tan(5\arctan(x))=\frac{5x-10x^3+x^5}{1-10x^2+5x^4},$$
...
A: Let $\displaystyle\arctan x=\theta\implies\tan\theta=x$
$$\tan(3\arctan x)=\tan3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$$
Put the values of $\tan\theta$
