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How can I show that the function $f(x) = \tan x$ is strictly increasing on $[-\pi/2, \pi/2]$ without using derivatives?

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Are you asking about the function $f(x) = \tan x$? If so, do you mean "how do I show $\tan x$ is strictly increasing on $[-\pi/2, \pi/2]$ without using derivatives"? $\tan x$ is not strictly increasing everywhere, since, for example, $\tan(-3\pi/4) = 1 > \tan(0) = 0$, but $-3\pi/4 < 0$. –  Henry T. Horton Feb 2 '13 at 21:09
    
Yes, that's what I meant, sorry. –  user58233 Feb 2 '13 at 21:18
    
Not to be nit-picky, but you want to say that $\tan x$ is strictly increasing on the open interval $(-\pi / 2,\pi / 2)$ since it is not defined on the endpoints of the closed interval. –  chharvey Feb 2 '13 at 21:34

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First understand that the tangent function is only strictly increasing in between its poles. So, within a region between adjacent poles, consider the difference formula (admittedly, from which the derivative may be derived):

$$\tan{x} - \tan{y} = \frac{\sin{x} \cos{y} - \cos{x} \sin{y}}{\cos{x} \cos{y}} = \frac{\sin{(x-y)}}{\cos{x} \cos{y}} $$

Note that, in between adjacent poles of the tangent function, the denominator is positive. Thus, when $x>y$, $\tan{x}>\tan{y}$.

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Go back to the basics.

Recall that in the unit circle trigonometry, $\tan \theta$ is represented by the slope (gradient) of the line which makes an angle of $\theta$ with respect to the x-axis.

This is clearly an increasing function on the domain $[ - \frac {\pi}{2}, \frac {\pi}{2} ]$.

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