Q.The function $f(x) = \tan x - x$:
(A) always increases
(B) always decreases
(C) never increases
(D) sometimes increases and sometimes decreases

The key idea is monotonicity of trigonometric functions — we check the sign of $f'(x)$.

Step 1: Differentiate $f(x) = \tan x - x$:

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

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

Why monotonicity? The core idea

To decide whether a function "always increases," "always decreases," or does something else, we look at its derivative. If $f'(x) \ge 0$ everywhere (and not identically zero on any interval), the function is non-decreasing — and if it's strictly positive except at isolated points, the function is strictly increasing. The same logic applies for decreasing with $f'(x) \le 0$.

Here, $f(x) = \tan x - x$. The derivative is straightforward, but we must be careful about the domain: $\tan x$ is undefined at $x = \frac{\pi}{2} + n\pi$, so we consider each continuous interval separately.

Step-by-step solution

1. Find the derivative

2. Simplify using a trigonometric identity Recall that $\sec^2 x = 1 + \tan^2 x$. Therefore:

3. Analyze the sign of $f'(x)$ The square of any real number is always non-negative. So:

Hence $f'(x) \ge 0$ on every interval of continuity.

4. Where is $f'(x) = 0$? $\tan^2 x = 0$ exactly when $\tan x = 0$, i.e., at $x = n\pi$ (integer $n$). These are isolated points — they do not form an interval. Between these points, $\tan^2 x > 0$. …

Method: Proving Strict Monotonicity Using a Trigonometric Identity

Apply this when a derivative expression can be simplified into a perfect square (or similarly always-nonnegative form) using a standard trig identity, to prove a function always increases or always decreases.

Steps

Step 1: Differentiate the function normally.

Apply the standard derivative rules — here $\frac{d}{dx}\tan x=\sec^2 x$ — to get the raw derivative expression.

Step 2: Rewrite the derivative using a known identity to reveal a squared (always non-negative) term.

Common identities like $\sec^2 x = 1+\tan^2 x$ let you simplify a derivative that doesn't look obviously signed into one that clearly is:

Step 3: Argue that a squared quantity is never negative.

Since any real number squared is $\ge 0$, the rewritten derivative is $\ge 0$ everywhere it is defined. …

Common Mistakes

Mistake 1: Thinking $f'(x)=0$ at some points means the function is constant there.

Why it's wrong: a zero derivative at isolated, separated points (not on a whole interval) does not make the function constant — it is flat only for an instant and continues increasing immediately afterward. Correct approach: check whether $f'(x)=0$ holds on an entire interval or only at scattered points; only the former implies a constant piece.

Mistake 2: Ignoring the domain restrictions of $\tan x$. …