Find the fifth expression of Taylor's for $\sin (\tan x)$ around $x=0$ 
Find the fifth expression of Taylor's for $\sin (\tan x)$ around $x=0$

My try:
$$\sin x=x+r_{1}(x), r_{1}(x)=o(x)$$ $$\tan x=x+\frac{x^3}{3}+\frac{2}{15}x^5+r_{2}(x), r_{2}(x)=o(x^5)$$ So: $$\sin \tan x=(x+\frac{x^3}{3}+\frac{2}{15}x^5)+r_{3}(x+\frac{x^3}{3}+\frac{2}{15}x^5+r_{2}(x)))=x+\frac{x^3}{3}+\frac{2}{15}x^5+r_{4}(x), r_{4}(x)=o(x)$$
That is why fifth expression of Taylor's is $\frac{2}{15}x^5$ for me. However Mathematica say that: $$\sin \tan x=x+\frac{x^3}{6}-\frac{x^5}{40}+o(x^5)$$ Where have I a mistake?

Hint:
Try starting $$\sin (\tan x)=\tan x - \frac{(\tan x)^3}{3!} + \frac{(\tan x)^5}{5!} +o((\tan x)^5)$$
And you must consider all terms that have $x^5$ in them, for example, $(\tan x)^3= (x+\frac{x^3}{3} +\frac{2}{15}x^5)^3$ have $(\frac{x^5}{3}+\frac{2x^5}{3})$
So, if you gather all the $x^5$ terms respectively, you'll find out that the coefficient of $x^5$ in the series is $$\frac{2}{15} -\frac{1}{6}(\frac{1}{3}+\frac{2}{3})+\frac{1}{5!}= \frac{-3}{120}=\frac{-1}{40}$$
