Q.Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Concept: Continuity At A Point — A function $f$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$. For trigonometric functions, continuity follows from the continuity of sine and cosine, and the domain restrictions of their reciprocals.

1. Cosine: $f(x) = \cos x$ is continuous for all real $x$, since $\lim_{x \to a} \cos x = \cos a$ for every $a \in \mathbb{R}$.

2. Cosecant: $\csc x = \frac{1}{\sin x}$. It is continuous wherever $\sin x \neq 0$, i.e., on $\mathbb{R} \setminus \{n\pi : n \in \mathbb{Z}\}$. At $x = n\pi$, it has vertical asymptotes (discontinuities).

3. Secant: $\sec x = \frac{1}{\cos x}$. Continuous on $\mathbb{R} \setminus \{(2n+1)\frac{\pi}{2} : n \in \mathbb{Z}\}$, where $\cos x = 0$ gives vertical asymptotes. …

The Core Idea: Continuity at a Point

Continuity of a function at a point $x = a$ means three things must hold:

1. The function is defined at $a$ (i.e., $f(a)$ exists).
2. The limit $\lim_{x \to a} f(x)$ exists.
3. The limit equals the function value: $\lim_{x \to a} f(x) = f(a)$.

For trigonometric functions, the key fact is that sine and cosine are continuous everywhere on $\mathbb{R}$. This is a standard result from analysis — you can prove it using the inequality $|\sin x - \sin a| \leq |x - a|$ and the squeeze theorem. Once we have that, the continuity of the other four functions follows from the algebra of continuous functions: the quotient of two continuous functions is continuous wherever the denominator is non-zero.

Let's examine each function in turn.

1. Cosine function: $f(x) = \cos x$

Cosine is continuous at every real number. There is no denominator, no square root, no restriction. The graph is a smooth wave with no breaks, jumps, or holes.

Why? For any $a \in \mathbb{R}$, we have:

This is a standard limit taught in Class 11/12. The proof uses the identity:

and the fact that $|\sin t| \leq |t|$, giving $|\cos x - \cos a| \leq |x - a|$, which forces continuity.

2. Cosecant function: $f(x) = \csc x = \frac{1}{\sin x}$

Cosecant is the reciprocal of sine. Since sine is continuous everywhere, the reciprocal $1/\sin x$ is continuous wherever $\sin x \neq 0$.

Where is $\sin x = 0$? At integer multiples of $\pi$: $x = n\pi$, $n \in \mathbb{Z}$.

At these points, $\csc x$ is undefined — you cannot divide by zero. So the function has vertical asymptotes there, not removable discontinuities.

Conclusion: $\csc x$ is continuous at every $x$ where $\sin x \neq 0$.

3. Secant function: $f(x) = \sec x = \frac{1}{\cos x}$

Secant is the reciprocal of cosine. By the same logic, it is continuous wherever $\cos x \neq 0$.

Where is $\cos x = 0$? At odd multiples of $\frac{\pi}{2}$: $x = \frac{(2n+1)\pi}{2}$, $n \in \mathbb{Z}$.

At these points, $\sec x$ is undefined. Everywhere else, it is continuous.

Conclusion: $\sec x$ is continuous at every $x$ where $\cos x \neq 0$.

4. Cotangent function: $f(x) = \cot x = \frac{\cos x}{\sin x}$ …

Method: Determining the Continuity Domain of a Reciprocal/Quotient Trigonometric Function

Use this whenever asked about the continuity of $\csc x$, $\sec x$, $\cot x$, or any function written as one continuous function divided by another — the domain of continuity is everywhere except where the denominator is zero.

Steps

Step 1: Rewrite the function as a quotient of known continuous functions

E.g. $\csc x = \dfrac{1}{\sin x}$, $\sec x = \dfrac{1}{\cos x}$, $\cot x = \dfrac{\cos x}{\sin x}$ — express in terms of $\sin x$ and $\cos x$, which are already known to be continuous on all of $\mathbb{R}$.

Step 2: Apply the quotient rule for continuity

The numerator and denominator being continuous is automatic here (they're $\sin x$/$\cos x$); the only condition left to check is the denominator being nonzero.

Step 3: Find every point where the denominator vanishes …

Common Mistakes

Mistake 1: Saying "$\csc x$ is discontinuous at $x = n\pi$"

Why it's wrong: $\csc x$ isn't even defined at those points (division by zero), so continuity — a property of a function AT a point in its domain — doesn't apply there at all; it's not a discontinuity, it's simply outside the domain. Correct approach: state the function's domain first ($\mathbb{R}$ minus the zeros of the denominator), then say it is continuous everywhere ON that domain.

Mistake 2: Mixing up which trig function vanishes where …