Answer
$ x-\dfrac{x^3}{2}+\dfrac{3x^5}{8}-\dfrac{5x^7}{16}+....$
Work Step by Step
The Taylor series for $\sin x $ can be defined as: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$ and the Taylor series for $\tan^{-1} x $ at $ x=0$ can be defined as: $\tan^{-1} x= x-\dfrac{x^3}{3}+\dfrac{ x^5}{5}-....$
We have
$\sin (\tan^{-1} x) = x-\dfrac{(\tan^{-1} x)^3}{3!}+\dfrac{ (\tan^{-1} x)^5}{5!}-....$
or, $=(x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-....) -\dfrac{1}{3!} (x-\dfrac{x^3}{6}+\dfrac{x^5}{5}-....)^3 +....$
or, $=x-\dfrac{x^3}{2}+\dfrac{3x^5}{8}-\dfrac{5x^7}{16}+....$
