I would like to check whether $f: \Bbb [-\pi/2, \pi/2] \in \Bbb R \rightarrow \Bbb R$ where $f(x) = \tan x$ is monotone increasing function.
Of course it looks fairly true in my bare eye with denoted in the printed version of textbook, however, want to show it from the ground.
Which approach is a valid sense to show the monotone property in this case and general case also?
On
The domain should be $(-\pi/2,\pi/2)$.
Since $\displaystyle f'(x)=\sec^2x>0$ for all $\displaystyle x\in\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$, $f( x)$ is (strictly) increasing.
Alternatively, for $\pi/2\le x\le y\le\pi/2$,
\begin{align} f(y)-f(x)&=\frac{\sin y}{\cos y}-\frac{\sin x}{\cos x}\\ &=\frac{\sin y\cos x-\cos y\sin x}{\cos y\cos x}\\ &=\frac{\sin(y-x)}{\cos y\cos x}\\ \end{align}
Note that $\displaystyle y-x\in(0,\pi)$. So, we have $\sin(y-x)>0$.
Since $\cos y\cos x>0$, $f(y)>f(x)$.
$f$ is (strictly) increasing.
HINT: $f'(x)=\frac{1}{\cos^2(x)}>0$ for all $x$ from the given interval we use the Quotient rule: $$\tan(x))'=\frac{\sin^2(x)+\cos^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}>0$$