Prove the following $\tan(nA)$ expansion I've figured out the approach. Writing the expansion of $(1 + x)^n$, then replacing $x$ with $i \tan (A)$.
Then separating real and imaginary part and $\tan(nA)$ will be equal to Im/Real.
But, after reaching $(1 + i \tan(A))^n$, I'm unable to convert it into De-Moivre's form from which I could proceed further. 
Prove that

$$\tan(nA) = \dfrac{\dbinom{n}1t - \dbinom{n}3 t^3 + \dbinom{n}5 t^5 \pm \cdots}{1 - \dbinom{n}2 t^2 + \dbinom{n}4 t^4 \pm \cdots}$$
  where $t= \tan(A)$.

 A: Generally
$$
\tan(\alpha+\beta+\gamma+\cdots) = \frac{e_1-e_3+e_5-e_7+\cdots}{e_0 -e_2+e_4-e_6+\cdots}\tag{1}
$$
where $e_k$ is the sum of all products of $k$ of the tangents $\tan\alpha,\tan\beta,\tan\gamma,\ldots\  {}$.  For example, if there are just four variables, $\alpha,\beta,\gamma,\delta$, then
$$
e_2 = \tan\alpha\tan\beta + \tan\alpha\tan\gamma + \tan\alpha\tan\delta + \tan\beta\tan\gamma+\tan\beta\tan\delta +\tan\gamma\tan\delta
$$
and
$$
e_3 = \tan\alpha\tan\beta\tan\gamma+\tan\alpha\tan\beta\tan\delta+\tan\alpha\tan\gamma\tan\delta+\tan\beta\tan\gamma\tan\delta,
$$
and so on.  And of course $e_0=1$ (except when there are $0$ variables, in which case $e_0=0$).
It's easy to prove $(1)$ by mathematical induction on the number of variables.
So if you want $\tan(n\alpha)$, it's just the case where all of the variables are the same variable, $\alpha$.  So for example, if there are four variables, then $e_2 = \dbinom 4 2 \tan\alpha\tan\alpha = 6\tan^2\alpha$ and $e_3 = \dbinom 4 3 \tan^3\alpha$, etc.
A: \begin{align}
(1 + i \tan(A))^n & = \left( 1 + i \dfrac{\sin(A)}{\cos(A)}\right)^n\\
& = \left(\dfrac{\cos(A) + i \sin(A)}{\cos(A)} \right)^n\\
& = \dfrac{e^{inA}}{\cos^n(A)}\\
& = \dfrac{\cos(nA) + i \sin(nA)}{\cos^n(A)}
\end{align}
Hence, the real part is $\dfrac{\cos(nA)}{\cos^n(A)}$ and the imaginary part is $\dfrac{\sin(nA)}{\cos^n(A)}$.
Now you should be able to finish it off by dividing the imaginary part by the real part.
