Prove that $\tan{nA}=\frac{\binom{n}{1}\tan{A}-\binom{n}{3}\tan^3{A}+\binom{n}{5}\tan^5{A}-\cdots}{\binom{n}{0}\tan^0{A}-\binom{n}{2}\tan^2{A}+\binom{n}{4}\tan^4{A}-\cdots}$
This is a question from the chapter permutations and combinations and I have no idea as to how to apply those concepts in this question and it would be great if I could get a hint...
Hint Perhaps this can help.
$$\left(\frac{1 + i \tan A}{1 - i \tan A}\right)^n=\left(\frac{\cos A + i \sin A}{\cos A - i \sin A}\right)^n=\left(\frac{e^{iA}}{e^{-iA}}\right)^n=\frac{\cos nA + i \sin nA}{\cos nA - i \sin nA}=\frac{1+i \tan nA}{1-i \tan nA}$$