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(A) Given the function $f(x) = 	an x - x$. To determine the nature of the function,we find its derivative with respect to $x$: $f'(x) = \frac{d}{dx}(

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always increasing

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(A) Given the function $f(x) = 	an x - x$. To determine the nature of the function,we find its derivative with respect to $x$: $f'(x) = \frac{d}{dx}(	an x - x) = \sec^2 x - 1$. Using the trigonometric identity $1 + 	an^2 x = \sec^2 x$,we have $\sec^2 x - 1 = 	an^2 x$. Thus,$f'(x) = 	an^2 x$. Since the square of any real number is always non-negative,$	an^2 x \geq 0$ for all $x$ in the domain. Therefore,$f'(x) \geq 0$,which implies that the function $f(x)$ is always increasing.

The function $f(x) = \tan x - x$ is:

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