If $f:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow R$ and $g:[-1,1] \rightarrow R$ be defined as

Solution:
$g(x)=\sqrt{1-x^{2}}$
$\Rightarrow x^{2} \geq 0, \forall x \in[-1,1]$
$\Rightarrow-x^{2} \leq 0, \forall x \in[-1,1]$
$\Rightarrow 1-x^{2} \leq 1, \forall x \in[-1,1]$
We know that $1-x^{2} \geq 0$
$\Rightarrow 0 \leq 1-x^{2} \leq 1$
$\Rightarrow$ Range of $g(x)=[0,1]$
So, $f:\left(\frac{-\pi}{2}, \frac{\pi}{2}\right) \rightarrow R$ and $g:[-1,1] \rightarrow[0,1]$
Computation of fog:
Clearly, the range of $g$ is a subset of the domain of $f$.
So, fog: $[-1,1] \rightarrow R$
$(f \circ g)(x)=f(g(x))$
$=f\left(\sqrt{1-\mathrm{x}^{2}}\right)$
$=\tan \sqrt{1-\mathrm{x}^{2}}$
Computation of $g o f$ :
Clearly, the range of $f$ is not a subset of the domain of $g$.
$\Rightarrow$ Domain $(g o f)=\{x \in$ domain of $f$ and $\mathrm{f}(\mathrm{x}) \in$ domain of $g\}$
$\Rightarrow$ Domain $($ gof $)=\left\{x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\right.$ and $\left.\tan x \in[-1,1]\right\}$
$\Rightarrow$ Domain $($ gof $)=\left\{\mathrm{x} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\right.$ and $\left.\mathrm{x} \in\left(\frac{-\pi}{4}, \frac{\pi}{4}\right)\right\}$
$\Rightarrow$ Domain $($ gof $)=\left\{\mathrm{x} \in\left(\frac{-\pi}{4}, \frac{\pi}{4}\right)\right\}$
Now, gof : $\left(\frac{-\pi}{4}, \frac{\pi}{4}\right) \rightarrow R$
So, $(g o f)(x)=g(f(x))$
$=g(\tan x)$
$=\sqrt{1-\tan ^{2} x}$