Reciprocal Trigonometric Functions
Every function has its mirror image. For sine, cosine, and tangent, those mirrors are cosecant, secant, and cotangent. These reciprocal functions appear in physics, engineering, and astronomy whenever quantities are inversely related. In this lesson you will learn their definitions, how to evaluate them, and where they are undefined.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
If $\sin \theta = 0$, then $\frac{1}{\sin \theta}$ is undefined because you cannot divide by zero. Where on the unit circle does $\sin \theta = 0$? And what does this tell you about where $\csc \theta$ is undefined?
For every basic trigonometric function there is a corresponding reciprocal function. A number and its reciprocal always have the same sign — so ASTC applies directly to the reciprocal functions too.
The reciprocal functions are defined by flipping the original function. Cosecant is the reciprocal of sine, secant is the reciprocal of cosine, and cotangent is the reciprocal of tangent. They are undefined wherever their denominator equals zero.
Key facts
- The definitions of $\csc \theta$, $\sec \theta$, and $\cot \theta$
- Where each reciprocal function is undefined
- The relationship between original and reciprocal functions
Concepts
- Why reciprocal functions have vertical asymptotes
- How the sign of a reciprocal matches the sign of the original
- The domain restrictions introduced by reciprocals
Skills
- Evaluate reciprocal trig functions from exact values
- State the domain of each reciprocal function
- Simplify expressions involving reciprocal functions
Because division by zero is undefined, each reciprocal function has restrictions on its domain. The signs follow directly from the original function via ASTC:
- $\csc \theta$ is positive where $\sin \theta$ is positive (Quadrants I and II)
- $\sec \theta$ is positive where $\cos \theta$ is positive (Quadrants I and IV)
- $\cot \theta$ is positive where $\tan \theta$ is positive (Quadrants I and III)
The range of cosecant and secant is $(-\infty, -1] \cup [1, \infty)$ because they are reciprocals of numbers in $[-1, 1]$ (excluding 0). The range of cotangent is all real numbers.
Three reciprocal definitions: $\csc \theta = \frac{1}{\sin \theta}$, $\sec \theta = \frac{1}{\cos \theta}$, $\cot \theta = \frac{\cos \theta}{\sin \theta}$; Signs follow ASTC — a reciprocal has the same sign as its original function
Pause — copy the three reciprocal definitions ($\csc\theta = 1/\sin\theta$; $\sec\theta = 1/\cos\theta$; $\cot\theta = \cos\theta/\sin\theta$) and the sign rule (same sign as the original function) into your book.
We just saw that $\sec\theta = 1/\cos\theta$ and the reciprocals share the ASTC sign of their originals. That raises a question: how do we actually compute a reciprocal value like $\sec\frac{\pi}{3}$ in practice? This card answers it → evaluate the original function first, then take the reciprocal: $\sec\frac{\pi}{3} = 1/\cos\frac{\pi}{3} = 1/(1/2) = 2$.
Find the exact value of $\sec \frac{\pi}{3}$.
To evaluate a reciprocal trig function: write it as $\frac{1}{\text{original}}$, then substitute the exact value; $\sec \frac{\pi}{3} = \frac{1}{\cos \frac{\pi}{3}} = \frac{1}{1/2} = 2$
Pause — copy the reciprocal evaluation method: write as $1/\text{original}$, substitute exact value, simplify — and the worked result $\sec\frac{\pi}{3} = 2$ into your book.
We just saw that reciprocals are $1/\text{original}$. That raises a question: since division by zero is undefined, at which angles does each reciprocal function fail to exist? This card answers it → $\csc\theta$ is undefined when $\sin\theta = 0$; $\sec\theta$ when $\cos\theta = 0$; $\cot\theta$ when $\sin\theta = 0$.
State the values of $\theta$ in $[0, 2\pi]$ for which $\cot \theta$ is undefined.
$\cot \theta$ is undefined when $\sin \theta = 0$, i.e. $\theta = 0, \pi, 2\pi, \ldots$; $\sec \theta$ is undefined when $\cos \theta = 0$, i.e. $\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots$
Pause — copy the undefined-angle rules: $\cot\theta$ undefined at $\theta = 0, \pi, 2\pi,\ldots$ (where $\sin\theta = 0$); $\sec\theta$ undefined at $\theta = \frac{\pi}{2}, \frac{3\pi}{2},\ldots$ (where $\cos\theta = 0$) into your book.
We just saw the domain restrictions of reciprocal functions. That raises a question: when a reciprocal appears inside a larger expression, how do we simplify — do we just substitute the definition? This card answers it → yes: replace with $1/\text{original}$, then cancel pairs like $\sin\theta \cdot \csc\theta = 1$.
Simplify $\sin \theta \cdot \csc \theta + \cos \theta \cdot \sec \theta$.
Key identity: $\sin \theta \cdot \csc \theta = 1$ and $\cos \theta \cdot \sec \theta = 1$ (where defined); When simplifying, replace reciprocals with $\frac{1}{\text{original}}$ and cancel
Pause — copy the two cancellation identities: $\sin\theta \cdot \csc\theta = 1$ and $\cos\theta \cdot \sec\theta = 1$, plus the simplification strategy (replace with $1/\text{original}$, then cancel) into your book.
Students often swap $\sec \theta$ and $\csc \theta$. Remember the pairings: secant goes with cosine, cosecant goes with sine. The "co-" in cosecant matches sine (no "co-"), while secant matches cosine (which has "co-").
While $\cot \theta = \frac{1}{\tan \theta}$ is correct, it is often more useful to write $\cot \theta = \frac{\cos \theta}{\sin \theta}$, especially when simplifying identities or finding where it is undefined.
When asked for the domain of $\sec \theta$, some students say "all real numbers" or only exclude $\frac{\pi}{2}$. They forget that every odd multiple of $\frac{\pi}{2}$ must be excluded. Use the general form $n \in \mathbb{Z}$.
Work these through step-by-step. Use exact values and state any domain restrictions.
Find the exact value of $\csc \frac{\pi}{6}$.
$\csc \frac{\pi}{6} = \frac{1}{\sin \frac{\pi}{6}} = \frac{1}{\frac{1}{2}} = 2$.
Find the exact value of $\sec \frac{\pi}{4}$.
$\sec \frac{\pi}{4} = \frac{1}{\cos \frac{\pi}{4}} = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.
Find the exact value of $\cot \frac{\pi}{3}$.
$\cot \frac{\pi}{3} = \frac{1}{\tan \frac{\pi}{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$.
Find the exact value of $\sec \frac{2\pi}{3}$.
QII: $\cos \frac{2\pi}{3} = -\frac{1}{2}$, so $\sec \frac{2\pi}{3} = -2$.
Find the exact value of $\csc \frac{3\pi}{4}$.
QII: $\sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2}$, so $\csc \frac{3\pi}{4} = \sqrt{2}$.
- $\csc \theta$
- $\sec \theta$
- $\cot \theta$
- undefined when $\sin \theta = 0$
- undefined when $\cos \theta = 0$
- undefined when $\sin \theta = 0$
Return to your original answer from Section 01. $\sin \theta = 0$ at the points where the unit circle crosses the $x$-axis: $\theta = 0, \pi, 2\pi$, and so on. At these angles, the $y$-coordinate is 0. Since $\csc \theta = \frac{1}{\sin \theta}$, dividing by zero means $\csc \theta$ is undefined at every integer multiple of $\pi$. These points become vertical asymptotes on the graph of $y = \csc \theta$.
How does your initial thinking compare? Did you identify all three points in $[0, 2\pi]$?
Exact values with ASTC
(a) Find the exact value of $\csc \frac{5\pi}{6}$. (b) Find the exact value of $\sec \frac{7\pi}{4}$. Show reference angles and ASTC reasoning.
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(a) $\frac{5\pi}{6}$ is in QII. Reference angle = $\frac{\pi}{6}$. $\sin \frac{\pi}{6} = \frac{1}{2}$. In QII, sine is positive, so $\sin \frac{5\pi}{6} = \frac{1}{2}$. Therefore $\csc \frac{5\pi}{6} = \mathbf{2}$.
(b) $\frac{7\pi}{4}$ is in QIV. Reference angle = $\frac{\pi}{4}$. $\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$. In QIV, cosine is positive, so $\cos \frac{7\pi}{4} = \frac{\sqrt{2}}{2}$. Therefore $\sec \frac{7\pi}{4} = \mathbf{\sqrt{2}}$.
Simplify to a single function
Simplify $\frac{\sec \theta}{\tan \theta}$ to a single trigonometric function. State any restrictions on $\theta$.
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Working:
Answer: $\mathbf{\csc \theta}$
Restrictions: $\cos \theta \neq 0$ and $\sin \theta \neq 0$, so $\theta \neq \frac{n\pi}{2}$ for any integer $n$.
Explain the range of cosecant
Explain why the range of $y = \csc \theta$ is $(-\infty, -1] \cup [1, \infty)$. Use the relationship between $\csc \theta$ and $\sin \theta$ in your explanation.
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Since $\csc \theta = \frac{1}{\sin \theta}$ and $\sin \theta \in [-1, 1]$ (excluding 0), the reciprocal of a number in $(-1, 1)$ (excluding 0) lies outside $(-1, 1)$.
When $0 < \sin \theta \leq 1$, we have $\csc \theta \geq 1$.
When $-1 \leq \sin \theta < 0$, we have $\csc \theta \leq -1$.
Therefore the range is $(-\infty, -1] \cup [1, \infty)$.
Key insight: the reciprocal function "flips" the interval. Values close to 0 become very large, while the endpoints $\pm 1$ stay fixed.
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