<s>[INST] Prove the following statements without using induction The  first statement: 
$$\tan(nx)=\frac{C^n_1\tan(x)-C^n_3\tan^3(x)+C^n_5\tan^5(x)-C^n_7\tan^7(x)+\dotsm}{1-C^n_2\tan^2(x)+C^n_4\tan^4(x)-C^n_6\tan^6(x)+\dotsm}$$
where 
$$C^n_k=\frac{n!}{k!(n-k)!}$$
The second statement:
$$\sin(x)\sin\Bigl(x+\frac{\pi}{n}\Bigr)\sin\Bigl(x+\frac{2\pi}{n}\Bigr)\ ...\sin\Bigl(x+\frac{(n-1)\pi}{n}\Bigr)=2^{1-n}\sin(nx)$$
The third statement: 
$$\DeclareMathOperator{\cotg}{cotg}\cotg(x)+\cotg\Bigl(x+\frac{\pi}{n}\Bigr)+\cotg\Bigl(x+\frac{2\pi}{n}\Bigr)+\dotsm+\cotg\Bigl(x+\frac{(n-1)\pi}{n}\Bigr)=n\cotg(nx)$$
 [/INST] Use Binomial & DeMoivre's Theorems
\begin{eqnarray*}
\cos(nx) &=& \operatorname{Re}(( \cos x+ i \sin x)^n) =\cos^n x - \binom{n}{2} \cos^{n-2} x \sin^2 x + \binom{n}{4} \cos^{n-4} x \sin^4 x - \cdots\\
\sin(nx) &=&\operatorname{Im}(( \cos x+ i \sin x)^n) =\binom{n}{1}\cos^{n-1} x \sin x- \binom{n}{3} \cos^{n-3} x \sin^3 x +  \binom{n}{5} \cos^{n-5} x \sin^5 x - \cdots
\end{eqnarray*}
Now 
\begin{eqnarray*}
\tan(nx) &=& \frac{\sin nx}{\cos{nx}} \\
&=& \frac{\binom{n}{1}\cos^{n-1} x \sin x- \binom{n}{3} \cos^{n-3} x \sin^3 x + \binom{n}{5} \cos^{n-5} x \sin^5 x - \cdots}{ \cos^n x - \binom{n}{2} \cos^{n-2} x \sin^2 x + \binom{n}{4} \cos^{n-4} x \sin^4 x - \cdots} \\
&=& \frac{\binom{n}{1} \tan x- \binom{n}{3}  \tan^3 x + \binom{n}{5}  \tan^5 x - \cdots}{ 1 - \binom{n}{2}  \tan^2 x + \binom{n}{4}  \tan^4 x - \cdots} \\
\end{eqnarray*}
</s>