$f(x)=x/(1-x^{2})$ .

(a) Using the derivative: $f'(x)(1-x^{2})^{2}=1+x^{2}>0$ implies that $f$ is strictly increasing. Alternatively, we can argue that a) $f$ is an odd function on $(-1,1)$ ($f(-x)=-f(x)$ ), so it is sufficient to show that it is increasing on $[0;1)$ , b) for $0\le x<1$ , the numerator of $f(x)$ is strictly increasing in $x$ , while the denominator of $f(x)$ is strictly decreasing in $x$ , where both remain non-negative. Hence, $f(x)$ is strictly increasing.

(b) We first check that $g(f(x))=x$ for all $x\in(-1,1)$ . We can parametrize $x\in(-1,1)$ as follows: $x=\tan\alpha,-\pi/4<\alpha<\pi/4$ . Note that $\cos2\alpha>0$ , so $\cos2\alpha=\sqrt{\cos^{2}2\alpha}$ . Then, $f(x)=\tan2\alpha/2$ , and $g(f(x))=\tan2\alpha/[1+\sqrt{1+\tan^{2}2\alpha}]=\sin2\alpha/[\cos2\alpha+1]=\tan\alpha=x$ . Now, we can either check directly that $f(g(y))=y$ for all $y\in\mathbb{R}$ , or use the following argument. We have shown in (a) that $f$ is strictly increasing, and, hence, injective. It is also continuous having the negative and positive infinities as its limits when $x$ goes to $-1$ and $+1$ , respectively. Therefore, $f$ is surjective, and bijective. There exists the inverse of $f$ , $f^{-1}$ , which is its left and right inverses. According to Exercise 5 of §2, Chapter 1, in this case every left or right inverse of $f$ is equal to $f^{-1}$ , implying that $g=f^{-1}$ is the left and right inverses of $f$ .