## Calculus (3rd Edition)

We know that $\tan x$ is continuous on the interval $(-\frac{\pi}{2},\frac{\pi}{2})$ and $\sin x$ is continuous everywhere. Moreover, the range of $\sin x$ is $[-1,1]\subset (-\frac{\pi}{2},\frac{\pi}{2})$. Then, by Theorem 5 of the composition functions, the function $$f(x)=\tan (\sin x)$$ is continuous.