Answer
$B(x)$ is continuous on $(-2,\frac{-\pi}{2})\cup(\frac{-\pi}{2},\frac{\pi}{2})\cup(\frac{\pi}{2},2)$.
Work Step by Step
$$B(x)=\frac{\tan x}{\sqrt{4-x^2}}$$
1) Find the domain of the numerator
The domain of $\tan x$ is $x\in R$ except where $x=\frac{\pi}{2}+k\pi (k\in Z)$.
According to Theorem 7, $\tan x$ is continuous on $R$ except where $x=\frac{\pi}{2}+k\pi (k\in Z)$.
2) Find the domain of the denominator
$\sqrt{4-x^2}$ is defined for $$4-x^2\geq0$$$$x^2\le4$$$$-2\le x\le2$$
Therefore, the domain of $\sqrt{4-x^2}$ is $x\in[-2,2]$.
According to Theorem 7, $\sqrt{4-x^2}$ is continuous on $[-2,2]$.
3) Combining the domain in 1) and 2), we see that $\frac{\pi}{2}$ and $\frac{-\pi}{2}$ are within $[-2,2]$.
Therefore, we must exclude them out of the continuity range.
So, $\tan x$ and $\sqrt{4-x^2}$ are both continuous on $[-2,\frac{-\pi}{2})\cup(\frac{-\pi}{2},\frac{\pi}{2})\cup(\frac{\pi}{2},2]$.
According to part 5) of Theorem 4, $B(x)$ is also continuous on $[-2,\frac{-\pi}{2})\cup(\frac{-\pi}{2},\frac{\pi}{2})\cup(\frac{\pi}{2},2]$ except where $$\sqrt{4-x^2}=0$$$$4-x^2=0$$$$x^2=4$$$$\left\{\begin{array} {c l}x=2\\x=-2\end{array}\right.$$
Therefore, $B(x)$ is continuous on $(-2,\frac{-\pi}{2})\cup(\frac{-\pi}{2},\frac{\pi}{2})\cup(\frac{\pi}{2},2)$.
