Q.The function $f(x) = \cot x$ is discontinuous on the set
(A) $\{x = n\pi : n \in \mathbb{Z}\}$
(B) $\{x = 2n\pi : n \in \mathbb{Z}\}$
(C) $\left\{x = (2n + 1)\dfrac{\pi}{2} : n \in \mathbb{Z}\right\}$
(D) $\left\{x = \dfrac{n\pi}{2} : n \in \mathbb{Z}\right\}$

The key idea is that $\cot x = \frac{\cos x}{\sin x}$ is undefined (and therefore discontinuous) wherever $\sin x = 0$.

Step 1: $\cot x$ is a quotient of two continuous functions, so it is continuous wherever the denominator is non-zero.

Step 2: $\sin x = 0$ when $x = n\pi$, where $n \in \mathbb{Z}$. …

To understand why $\cot x$ is discontinuous at those points, we need to recall what discontinuity means for a function like this. A function is discontinuous at a point if it is not defined there, or if its limit does not exist or does not match the function value. For $\cot x$, the trouble comes entirely from division by zero.

1. Rewrite $\cot x$ in terms of sine and cosine.

   $\cot x = \frac{\cos x}{\sin x}$. This is the most useful form for analysing continuity because it shows the function is a quotient of two continuous functions — $\cos x$ and $\sin x$ are continuous everywhere. The only possible points of discontinuity are where the denominator $\sin x = 0$, because division by zero is undefined.

2. Find where $\sin x = 0$.

   From trigonometry, $\sin x = 0$ when $x = n\pi$, where $n$ is any integer ($n \in \mathbb{Z}$). So the set of problematic points is $\{x = n\pi : n \in \mathbb{Z}\}$.

3. Check what happens at these points.

   At $x = n\pi$, $\sin x = 0$ and $\cos x = \pm 1$ (specifically, $\cos(n\pi) = (-1)^n$). So $\cot x$ is not defined — you cannot divide by zero. Since the function does not exist at these points, it is certainly discontinuous there. Moreover, the left-hand and right-hand limits tend to $\pm \infty$ (the function blows up), so the discontinuity is not removable.

4. Compare with the given options.

   • Option (A): $\{x = n\pi\}$ — exactly matches our set.
   • Option (B): $\{x = 2n\pi\}$ — only half the points (where $\sin x = 0$ and $\cos x = 1$), misses $x = \pi, 3\pi, \dots$ …

Method: Finding Discontinuities of $\tan x$, $\cot x$, $\sec x$, $\csc x$

Any trigonometric ratio built as a quotient of sine and cosine is continuous exactly where its own denominator is non-zero — this method finds that set systematically.

Steps

Step 1: Rewrite the function purely in terms of $\sin x$ and $\cos x$.

Step 2: Note that $\sin x$ and $\cos x$ are continuous for every real $x$, so the only possible discontinuities of the ratio come from the denominator becoming zero.

Step 3: Solve for where that specific denominator is zero, using the correct trig identity. …

Common Mistakes

Mistake 1: Confusing the zeros of $\sin x$ (where $\cot x$ is undefined) with the zeros of $\cos x$ (where $\tan x$ is undefined).

Why it's wrong: $\cot x = \cos x/\sin x$ blows up where the denominator $\sin x=0$, i.e. at $x=n\pi$ — not at $x=(2n+1)\frac{\pi}{2}$, which is where $\cos x=0$ and $\cot x$ is actually perfectly defined (equal to $0$).

Correct approach: always identify which trig function is in the denominator of the given ratio before solving for its zero — don't rely on memorized 'trig blows up here' facts without checking which function is which.

Mistake 2: Picking option (D), $\{x=\frac{n\pi}{2}\}$, because it 'looks like it includes everything' and seems like the safe, inclusive choice. …