Exercise 4.7.8 (Tangent function is monotone, differentiable and invertible)

Let tan $:(-\pi ∕2,\pi ∕2)\to R$ be the tangent function $\tan <!--\; FUNCTION\; APPLICATION\; -->(x):=\sin <!--\; FUNCTION\; APPLICATION\; -->(x)∕\cos <!--\; FUNCTION\; APPLICATION\; -->(x)$. Show that tan is differentiable and monotone increasing, with $\frac{d}{\mathit{dx}}\tan <!--\; FUNCTION\; APPLICATION\; -->(x)=1+\tan <!--\; FUNCTION\; APPLICATION\; -->{(x)}^2$, and that $\underset{x\to \pi ∕2}{\lim <!--\; FUNCTION\; APPLICATION\; -->}\tan <!--\; FUNCTION\; APPLICATION\; -->(x)=+\infty$ and $\underset{x\to -\pi ∕2}{\lim <!--\; FUNCTION\; APPLICATION\; -->}\tan <!--\; FUNCTION\; APPLICATION\; -->(x)=-\infty$. Conclude that tan is in fact a bijection from $(-\pi ∕2,\pi ∕2)\to R$, and thus has an inverse function ${\tan <!--\; FUNCTION\; APPLICATION\; -->}^{-1}:R\to (-\pi ∕2,\pi ∕2)$ (this function is called the arctangent function). Show that ${\tan <!--\; FUNCTION\; APPLICATION\; -->}^{-1}$ is differentiable and $\frac{d}{\mathit{dx}}{\tan <!--\; FUNCTION\; APPLICATION\; -->}^{-1}(x)=\frac{1}{1+x^2}$.