Explanation

(b)
Given, $f(x) = \tan\sqrt{\frac{\pi}{9} - x^{2}}$
For $f(x)$ to be defined $\frac{\pi^{2}}{9} - x^{2} \geq 0$
$\Rightarrow \ \ \ \ x^{2} \leq \frac{\pi^{2}}{9} \Rightarrow - \frac{\pi}{3} \leq 3 \leq \frac{\pi}{3}$
$\therefore$ Domain of $f = \left\lbrack - \frac{\pi}{3},\ \frac{\pi}{3} \right\rbrack$
The greatest value of $f(x) = \tan\sqrt{\frac{\pi^{2}}{9}} - 0$, when $x = 0$
And the least value of $f(x) = \tan\sqrt{\frac{\pi^{2}}{9} - \frac{\pi^{2}}{9}}$, when $x = \frac{\pi}{3}$
$\therefore$ The greatest value of $f(x) = \sqrt{3}$ and the least value of $f(x) = 0$
$\therefore$ Range of $f = \lbrack 0,\ \sqrt{3}\rbrack$.