If $\tan(nx)$ is expanded in powers of $\tan(x)$ then what are the constant term and coefficient of $\tan(x)$ in the expansion?
My try I tried solving by comparing with the general formula for $\tan(nx)$ which is given in terms of $\tan(x)$ , $\tan^n(x)$ but did not get the correct answer.
Could someone please help me with this?
For real $x$ and $n \in \mathbb{Z}_{+}$, we have
$$\cos(nx)(1 + i \tan(nx)) = e^{inx} = (e^{ix})^n = (\cos(x)(1 + i\tan(x))^n$$ Compare real and imaginary parts on both side, we have
$$ \tan(nx) = \frac{\Im\rm(LHS)}{\Re\rm(LHS)} =\frac{\Im\rm(RHS)}{\Re\rm(RHS)} = \frac{\Im\left((1+i\tan x)^n\right)}{\Re\left((1+i\tan x)^n\right)} $$ This leads to $$\begin{align}\tan(nx) &= \frac{ \sum\limits_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor} (-1)^k \binom{n}{2k+1}\tan(x)^{2k+1} }{ \sum\limits_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor} (-1)^k\binom{n}{2k} \tan(x)^{2k} } = \frac{n \tan(x) + O(\tan(x)^3)}{1 + O(\tan(x)^2)}\\ &= 0 + n\tan(x) + O(\tan(x)^3) \end{align} $$ So the constant term is $0$ and the coefficient in front of $\tan(x)$ is $n$.
If $t = \tan(x)$, then you're expanding $\tan(n \arctan(t))$ in a Taylor series about $t=0$. Since $\arctan(t) = t + O(t^3)$ and $\tan(x) = x + O(t^3)$, we get $$ \tan(n \arctan(t)) = n t + O(t^3)$$
BTW, the next term is $$ \frac{n^3-n}{3} t^3$$
