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.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
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 the function $\csc \theta$ is undefined?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: sin(x) = 0.5 has only one solution: x = 30°.
Right: Trigonometric equations have infinitely many solutions due to periodicity. Within 0°–360°, sin(x) = 0.5 has solutions at x = 30° and x = 150°.
📚 Core Content
For every basic trigonometric function, there is a corresponding reciprocal function:
| Original | Reciprocal | Definition |
|---|---|---|
| $\sin \theta$ | $\csc \theta$ (cosecant) | $\frac{1}{\sin \theta}$ |
| $\cos \theta$ | $\sec \theta$ (secant) | $\frac{1}{\cos \theta}$ |
| $\tan \theta$ | $\cot \theta$ (cotangent) | $\frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}$ |
Because division by zero is undefined, each reciprocal function has restrictions on its domain:
A number and its reciprocal always have the same sign. This means ASTC applies directly to the reciprocal functions as well:
🧮 Worked Examples
🧪 Activities
1 $\csc \frac{\pi}{6}$
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2 $\sec \frac{\pi}{4}$
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3 $\cot \frac{\pi}{3}$
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4 $\sec \frac{2\pi}{3}$
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5 $\csc \frac{3\pi}{4}$
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1 $y = \sec \theta$
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2 $y = \csc \theta$
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3 $y = \cot \theta$
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Earlier you were asked: Where on the unit circle does $\sin \theta = 0$, and what does this tell you about $\csc \theta$?
$\sin \theta = 0$ 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 $y = \csc \theta$.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
8. (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. 3 MARKS
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Answer in your workbook.
9. Simplify the expression $\frac{\sec \theta}{\tan \theta}$ to a single trigonometric function. State any restrictions on $\theta$. 3 MARKS
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10. 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. 3 MARKS
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Answer in your workbook.
1. $\csc \frac{\pi}{6} = \frac{1}{\frac{1}{2}} = 2$
2. $\sec \frac{\pi}{4} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}$
3. $\cot \frac{\pi}{3} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$
4. $\sec \frac{2\pi}{3} = \frac{1}{-\frac{1}{2}} = -2$ (QII)
5. $\csc \frac{3\pi}{4} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}$ (QII)
1. $\sec \theta$ undefined at $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$
2. $\csc \theta$ undefined at $\theta = 0, \pi, 2\pi$
3. $\cot \theta$ undefined at $\theta = 0, \pi, 2\pi$
1. A — $\csc \theta = \frac{1}{\sin \theta}$.
2. A — $\sec \theta = \frac{1}{\cos \theta}$.
3. A — $\csc \theta = \frac{1}{\frac{1}{2}} = 2$.
4. A — $\cot \theta = \frac{\cos \theta}{\sin \theta}$.
5. A — $\sec \theta = \frac{1}{\frac{3}{5}} = \frac{5}{3}$.
Q8 (3 marks): (a) QII, reference $\frac{\pi}{6}$, $\sin = \frac{1}{2}$, so $\csc = 2$ [1.5]. (b) QIV, reference $\frac{\pi}{4}$, $\cos = \frac{\sqrt{2}}{2}$, so $\sec = \sqrt{2}$ [1.5].
Q9 (3 marks): $\frac{\sec \theta}{\tan \theta} = \frac{\frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}} = \frac{1}{\sin \theta} = \csc \theta$ [2]. Restrictions: $\cos \theta \neq 0$ and $\sin \theta \neq 0$, so $\theta \neq \frac{n\pi}{2}$ [1].
Q10 (3 marks): Since $\csc \theta = \frac{1}{\sin \theta}$ [1], and $\sin \theta \in [-1, 1]$ [0.5], the reciprocal of a number in $(-1, 1)$ (excluding 0) lies outside $(-1, 1)$ [1]. Therefore the range is $(-\infty, -1] \cup [1, \infty)$ [0.5].
Scale the platforms using your knowledge of sec, cosec and cot functions. Pool: lessons 1–6.
Tick when you've finished all activities and checked your answers.