The function $f(x) = tanx - x$

The function $f(x) = \tan x - x$ is defined on the interval $[0, \pi/2)$.

To determine if the function is increasing or decreasing, find the derivative:

$f'(x) = \sec^2 x - 1$

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Using the identity $\sec^2 x = 1 + \tan^2 x$:

$f'(x) = 1 + \tan^2 x - 1$

$f'(x) = \tan^2 x$

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For any real number, the square is non-negative:

$\tan^2 x \geq 0$ for all $x$ in $[0, \pi/2)$

$\tan^2 x = 0$ only when $x = 0$

$\tan^2 x > 0$ for all $x$ in $(0, \pi/2)$

Therefore $f'(x) \geq 0$ on $[0, \pi/2)$.

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Since $f'(x) > 0$ for $x \in (0, \pi/2)$, the function $f(x) = \tan x - x$ is increasing on $[0, \pi/2)$.