#### Explain solution for RD Sharma Class 12 Chapter Functions Exercise 2.3 Question 9 maths textbook solution.

\begin{aligned} &(f \circ g)(x)=\tan \left(\sqrt{1-x^{2}}\right) \\ &(g \circ f)(x)=\tan \left(\sqrt{1-x^{2}}\right) \end{aligned}

Given : Here given that

\begin{aligned} &f:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow R \text { and }\\ &g:[-1,1] \rightarrow R \text { defined as } f(x)=\tan x \text { and } g(x)=\sqrt{1-x^{2}} \end{aligned}

Hint : Since $\in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), y \in(-\infty, \infty)$

Range of $f\; \; \subset$ domain of $g = \left [ -1,1 \right ]$

Solution :

First, we compute $(f\; \circ \; g)(x)$

We know that

Let $y=f(x)$

$\begin{array}{ll} \Rightarrow \quad y=\tan \tan x \\ \Rightarrow \quad x=y ; x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \\ \Rightarrow \quad(f \circ g)(x)=f(g(x)) \\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;=f\left(\sqrt{1-x^{2}}\right) \\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; =\tan \left(\sqrt{1-x^{2}}\right) \end{array}$

Again we will compute $(g\; \circ \; f)(x)$

\begin{aligned} (g \circ f)(x) &=g(f(x)) \\ &=g(\tan x) \\ g \circ f(x)=& \tan \left(\sqrt{1-x^{2}}\right) \end{aligned}