If $f(x) = a(\tan x - \cot x)$, where $a > 0$, then find whether $f(x)$ is increasing or decreasing function in its domain.

If f(x) = a(tan x - cot x), where a > 0, then find whet — CBSE Class 12 Mathematics 2023 PYQ

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f(x) = a(\tan x - \cot x)

CBSECBSE Class XII Board 2023Subjective· 2mImportance★★★★★

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Concept understanding — Monotonicity of Trigonometric Functions

Monotonicity of Trigonometric Functions

The trigonometric functions rise and fall in a repeating pattern, so unlike a polynomial they are not monotonic over the whole real line — but on each piece of a period they are strictly increasing or strictly decreasing. Derivatives pin down exactly which piece is which.

The Idea

Picture the unit circle. As the angle $x$ grows, $\sin x$ (the height) climbs from $-1$ up to $1$ and back down, while $\cos x$ (the horizontal coordinate) does the same shifted by a quarter turn. Because the motion reverses at the top and bottom, each function alternates between increasing and decreasing stretches.

Sine

$\dfrac{d}{dx}\sin x = \cos x$ , so the sign of $\cos x$ decides the monotonicity of $\sin x$ :

$\cos x > 0$ on $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ , so $\sin x$ is strictly increasing there.
$\cos x < 0$ on $\left(\tfrac{\pi}{2}, \tfrac{3\pi}{2}\right)$ , so $\sin x$ is strictly decreasing there.

This pattern repeats every $2\pi$ .

Cosine

$\dfrac{d}{dx}\cos x = -\sin x$ , so the sign of $-\sin x$ governs $\cos x$ :

On $(0, \pi)$ , $\sin x > 0$ , hence $-\sin x < 0$ : $\cos x$ is strictly decreasing.
On $(\pi, 2\pi)$ , $\sin x < 0$ , hence $-\sin x > 0$ : $\cos x$ is strictly increasing.

Tangent

$\dfrac{d}{dx}\tan x = \sec^2 x > 0$ wherever it is defined. So $\tan x$ is strictly increasing on every interval $\left(-\tfrac{\pi}{2} + n\pi,\ \tfrac{\pi}{2} + n\pi\right)$ between its vertical asymptotes — but it does not carry that increase across an asymptote, so it is not monotonic on the whole line. …

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The function $f(x) = a(\tan x - \cot x)$ is increasing in every interval of its domain because its derivative $f'(x) = a \sec^2 x + a \csc^2 x$ is always positive for $a > 0$ .

Why monotonicity matters — and how to check it

When we ask whether a function is increasing or decreasing, we are really asking: as $x$ grows, does $f(x)$ go up, go down, or stay flat? The cleanest way to settle this is to look at the derivative. If $f'(x) > 0$ everywhere in the domain, the function is strictly increasing; if $f'(x) < 0$ , it is strictly decreasing.

The catch here is the domain. Both $\tan x$ and $\cot x$ blow up at certain points — $\tan x$ is undefined at $x = \frac{\pi}{2} + n\pi$ , and $\cot x$ is undefined at $x = n\pi$ . So the domain of $f$ is all real numbers except integer multiples of $\frac{\pi}{2}$ :

x \in \mathbb{R} \setminus \left\{ \frac{n\pi}{2} : n \in \mathbb{Z} \right\}

Within each continuous interval between these excluded points, we can differentiate freely.

Step-by-step solution

1. Rewrite the function in a simpler form

We have $f(x) = a(\tan x - \cot x)$ . Recall that $\cot x = \frac{1}{\tan x}$ , so:

f(x) = a\left( \tan x - \frac{1}{\tan x} \right)

This isn't strictly necessary, but it hints that the function might simplify further. Let's instead use the identity $\tan x - \cot x = \frac{\sin x}{\cos x} - \frac{\cos x}{\sin x}$ .

2. Combine into a single fraction

\tan x - \cot x = \frac{\sin^2 x - \cos^2 x}{\sin x \cos x} = -\frac{\cos 2x}{\frac{1}{2}\sin 2x} = -2\cot 2x

So $f(x) = -2a \cot 2x$ . This is a neat simplification — but we don't actually need it for the derivative approach. Let's proceed with the original form.

3. Differentiate $f(x)$

f'(x) = a \cdot \frac{d}{dx}(\tan x - \cot x) = a(\sec^2 x + \csc^2 x)

Why plus? Because $\frac{d}{dx}(\cot x) = -\csc^2 x$ , so subtracting $\cot x$ gives $-(-\csc^2 x) = +\csc^2 x$ .

4. Examine the sign of $f'(x)$

We know $a > 0$ by the problem statement. Now, $\sec^2 x = \frac{1}{\cos^2 x} \geq 0$ and $\csc^2 x = \frac{1}{\sin^2 x} \geq 0$ . At any point where both $\sin x$ and $\cos x$ are defined and non-zero (which is exactly the domain of $f$ ), both squares are strictly positive:

\sec^2 x > 0 \quad \text{and} \quad \csc^2 x > 0

Therefore: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. The function 'f' given by f(x) = log(cos x) is strictly decreasing on$
(a) (0, π/2)
(b) (π/2, π)
(c) (3π/2, 2π)
(d) None of the above
›Reveal solutionSolution

$f$ is decreasing exactly where $f'(x)<0$ AND the function is actually defined there (needs $\cos x>0$ ).

$f(x) = \log(\cos x)$ , so $f'(x) = \dfrac{-\sin x}{\cos x} = -\tan x$ .

For $f$ to be strictly decreasing, we need $f'(x) < 0 \Rightarrow \tan x > 0$ .

But we also need $\cos x > 0$ for $\log(\cos x)$ to even be defined (real-valued).

…

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CBSE 2025Set ANNUAL1 markMCQ
$\left(0,\dfrac{\pi}{2}\right)$
(a) $\sin x$
(b) $\cos x$
(c) $\tan x$
(d) $\sin 2x$
›Reveal solutionSolution

A function is strictly decreasing on an interval where its derivative is negative there.

On $\left(0,\dfrac{\pi}{2}\right)$ : $\dfrac{d}{dx}(\sin x)=\cos x>0$ (increasing); $\dfrac{d}{dx}(\cos x)=-\sin x<0$ (decreasing); $\dfrac{d}{dx}(\tan x)=\sec^2 x>0$ (increasing); $\dfrac{d}{dx}(\sin 2x)=2\cos 2x$ , which changes sign in this interval (not strictly monotonic throughout).

…

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CBSE 2025Set ANNUAL1 markMCQ
$f(x)=\sin x$
(a) $\left(0,\dfrac{\pi}{2}\right)$
(b) $\left(\dfrac{\pi}{2},\pi\right)$
(c) $(0,\pi)$
(d) $(0,2\pi)$
›Reveal solutionSolution

$\sin x$ increases exactly where $\cos x>0$ ; among the options that is $(0,\pi/2)$ .

Let $f(x)=\sin x$ . Then $f'(x)=\cos x$ .

$f$ is increasing on an interval where $f'(x)=\cos x>0$ .

On $(0,\pi/2)$ : $\cos x>0$ throughout, so $f$ is increasing here.

On $(\pi/2,\pi)$ : $\cos x<0$ , so $f$ is decreasing.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The function f(x) = sin x is increasing in -$
(i) (π/2, π)
(ii) (π, 3π/2)
(iii) (0, π)
(iv) (−π/2, π/2)
›Reveal solutionSolution
$\sin x$ increases exactly where $\cos x>0$ ; check each given interval against that sign.

$f(x)=\sin x \Rightarrow f'(x)=\cos x$ . $f$ is increasing on an interval where $\cos x>0$ .

$(\pi/2,\pi)$ : $\cos x<0$ here (second quadrant) — decreasing, not this one.

$(\pi,3\pi/2)$ : $\cos x<0$ here (third quadrant) — decreasing, not this one. …
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CBSE 2024Set ANNUAL1 markMCQ
$f(x)=\sin x$
(a) f(x) is a constant function
(b) f(x) is an increasing function
(c) f(x) is a decreasing function
(d) f(x) is neither increasing nor decreasing function
›Reveal solutionSolution

From the graph, $\sin x$ rises on $(0,\pi/2)$ and falls on $(\pi/2,\pi)$ , so it is not monotonic on the whole interval $(0,\pi)$ .

From the given graph of $f(x) = \sin x$ : on $(0, \pi/2)$ the curve rises from $0$ to its maximum value $1$ at $x=\pi/2$ , so $f'(x) = \cos x > 0$ there and $f$ is increasing on $(0,\pi/2)$ .

On $(\pi/2, \pi)$ the curve falls from $1$ back down to $0$ at $x=\pi$ , so $f'(x) = \cos x < 0$ there and $f$ is decreasing on $(\pi/2,\pi)$ .

…

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CBSE 2023Set ANNUAL1 mark
$x$
›Reveal solutionSolution

Since $f'(x)=\cos x - 1 \le 0$ for every real $x$ , the function never increases; it is monotonically decreasing throughout $\mathbb{R}$ .

Differentiate: $f'(x) = \cos x - 1$ .

Since $\cos x \le 1$ for all real $x$ , we get $f'(x) = \cos x - 1 \le 0$ for every $x \in \mathbb{R}$ , with equality only at the isolated points $x = 2n\pi$ , $n \in \mathbb{Z}$ .

…

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CBSE 2020Set ANNUAL1 markMCQ
$f(x) = \log(\sin x)$
(a) $\left(0, \frac{\pi}{2}\right)$
(b) $\left(\frac{\pi}{2}, \pi\right)$
(c) $(0, \pi)$
(d) None of these
›Reveal solutionSolution

$f'(x)=\cot x$ , which is negative on $\left(\frac{\pi}{2},\pi\right)$ , making $f$ decreasing there.

$f'(x) = \dfrac{\cos x}{\sin x} = \cot x$

On $\left(0,\dfrac{\pi}{2}\right)$ , $\cot x>0$ so $f$ is increasing. …

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Q.If f(x)=a(tan⁡x−cot⁡x)f(x) = a(\tan x - \cot x)f(x)=a(tanx−cotx), where a>0a > 0a>0, then find whether f(x)f(x)f(x) is increasing or decreasing function in its domain.

Why monotonicity matters — and how to check it

Step-by-step solution

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Q.If $f(x) = a(\tan x - \cot x)$ , where $a > 0$ , then find whether $f(x)$ is increasing or decreasing function in its domain.

Unlock everything free for 14 days