Which of the following functions are increasing on $x \in \left(0, \frac{\pi}{2}\right)$?

(A) $f(x) = \sin x$

(B) $f(x) = \cos x$

(C) $f(x) = \tan x$

(D) $f(x) = \cos 3x$

Choose the correct answer from the options given below:

A function is increasing when its derivative is positive throughout the interval.

The interval is $x \in \left(0, \frac{\pi}{2}\right)$.

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For $f(x) = \sin x$:

$f'(x) = \cos x$

For $x \in \left(0, \frac{\pi}{2}\right)$, $\cos x > 0$.

Therefore, $f(x) = \sin x$ is increasing on the interval.

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For $f(x) = \cos x$:

$f'(x) = -\sin x$

For $x \in \left(0, \frac{\pi}{2}\right)$, $\sin x > 0$, which means $-\sin x < 0$.

Therefore, $f(x) = \cos x$ is decreasing on the interval.

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For $f(x) = \tan x$:

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

For $x \in \left(0, \frac{\pi}{2}\right)$, $\sec^2 x > 0$ since any squared term is positive.

Therefore, $f(x) = \tan x$ is increasing on the interval.

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For $f(x) = \cos 3x$:

$f'(x) = -3\sin 3x$

When $x \in \left(0, \frac{\pi}{2}\right)$, then $3x \in \left(0, \frac{3\pi}{2}\right)$.

The function $\sin 3x$ is positive when $3x \in (0, \pi)$ and negative when $3x \in \left(\pi, \frac{3\pi}{2}\right)$.

This corresponds to $x \in \left(0, \frac{\pi}{3}\right)$ and $x \in \left(\frac{\pi}{3}, \frac{\pi}{2}\right)$ respectively.

Since $f'(x) = -3\sin 3x$ changes sign on the interval, the function is not consistently increasing.

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Only $f(x) = \sin x$ and $f(x) = \tan x$ are increasing throughout the entire interval $\left(0, \frac{\pi}{2}\right)$.

The answer is (A) and (C) only.