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Year 11 Specialist (Unit 1 & 2) Trigonometry and functions

Reciprocals of trigonometric functions

20 practice questions 0 video lessons Theory + worked examples

Get to grips with the reciprocal trigonometric functionssecant, cosecant and cotangent — for Year 11 Specialist Mathematics in Queensland (QCAA). Each one is simply a familiar ratio turned upside down: secant reciprocates cosine, cosecant reciprocates sine, and cotangent reciprocates tangent.

You will learn to define and use these functions, work out simplified exact values at the special angles, and recognise where each function is undefined — the groundwork for the Pythagorean identities and trigonometric modelling later in the course.

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Theory

The reciprocal trigonometric functionssecant, cosecant and cotangent — are the reciprocals of cosine, sine and tangent in Year 11 Specialist Mathematics (QCAA, Queensland). This page defines them, shows how to read off simplified exact values, and where each one is undefined.

The three reciprocal trigonometric functions are built by turning each familiar ratio upside down. Secant is the reciprocal of cosine, cosecant is the reciprocal of sine, and cotangent is the reciprocal of tangent.

In symbols, \(\sec x=\dfrac{1}{\cos x}\), \(\operatorname{cosec} x=\dfrac{1}{\sin x}\) and \(\cot x=\dfrac{1}{\tan x}=\dfrac{\cos x}{\sin x}\). Australian courses write cosecant as cosec (the same function is written \(\csc\) elsewhere).

Because you cannot divide by zero, each reciprocal function is undefined wherever its base function equals zero: \(\sec x\) fails where \(\cos x=0\), while \(\operatorname{cosec} x\) and \(\cot x\) both fail where \(\sin x=0\).

To evaluate one at a special angle, find the exact value of the base ratio first, then take its reciprocal and rationalise any surd. Every answer stays exact — surds and \(\pi\) are kept, never turned into decimals.

Unit circle and the reciprocal ratios A unit circle with a radius drawn to a point P at 60 degrees. The horizontal coordinate of P is cos and the vertical coordinate is sin, so sec is one over cos and cosec is one over sin. x y P θ cos sin
On the unit circle the point \(P\) has coordinates \((\cos\theta,\sin\theta)\), so \(\sec\theta=\dfrac{1}{\cos\theta}\) and \(\operatorname{cosec}\theta=\dfrac{1}{\sin\theta}\).
Special right triangles for exact values Left: a 45-45-90 triangle with two sides of length 1 and hypotenuse root 2. Right: a 30-60-90 triangle with sides 1, root 3 and hypotenuse 2. 1 1 √2 45° 45° √3 1 2 30° 60°
The special triangles give the base ratios: e.g. \(\cos\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}\), so \(\sec\dfrac{\pi}{6}=\dfrac{2\sqrt{3}}{3}\).

The three definitions, each undefined where its denominator is zero:

\[ \sec x=\dfrac{1}{\cos x}, \qquad \operatorname{cosec} x=\dfrac{1}{\sin x}, \qquad \cot x=\dfrac{1}{\tan x}=\dfrac{\cos x}{\sin x} \]
secx=1cosx

The simplified exact values at the first-quadrant special angles (read a base ratio, then take its reciprocal):

\(x\)\(0\)\(\dfrac{\pi}{6}\)\(\dfrac{\pi}{4}\)\(\dfrac{\pi}{3}\)\(\dfrac{\pi}{2}\)
\(\sec x\)\(1\)\(\dfrac{2\sqrt{3}}{3}\)\(\sqrt{2}\)\(2\)undefined
\(\operatorname{cosec} x\)undefined\(2\)\(\sqrt{2}\)\(\dfrac{2\sqrt{3}}{3}\)\(1\)
\(\cot x\)undefined\(\sqrt{3}\)\(1\)\(\dfrac{\sqrt{3}}{3}\)\(0\)
Reciprocal, not inverse. \(\sec x\) means \(\dfrac{1}{\cos x}\); it is not \(\cos^{-1}x\) and it is not \(\dfrac{1}{x}\). The angle stays inside the trig function.

How to evaluate a reciprocal function

  1. Identify the base function: secant pairs with cosine, cosecant with sine, cotangent with tangent (or cosine over sine).
  2. Evaluate the base ratio at the angle, using the special triangles or the unit circle, and keep the correct sign for the quadrant.
  3. Reciprocate: take one over that value; if the base ratio is \(0\), the reciprocal is undefined.
  4. Rationalise any surd denominator so the exact value is in simplest form.
Example 1 — whole-number value
Find the exact value of \(\sec\dfrac{\pi}{3}\).
Solution

Secant is the reciprocal of cosine; evaluate the cosine, then flip it:

\(\sec\dfrac{\pi}{3}\)\(=\)\(\dfrac{1}{\cos\dfrac{\pi}{3}}\)
\(\cos\dfrac{\pi}{3}\)\(=\)\(\dfrac{1}{2}\)
\(=\)\(\dfrac{1}{\dfrac{1}{2}}\)
\(=\)\(2\)

\(\sec\dfrac{\pi}{3}=2\).

Example 2 — rationalise a surd
Find the exact value of \(\operatorname{cosec}\dfrac{\pi}{3}\).
Solution

Cosecant is the reciprocal of sine; flip, then rationalise the surd denominator:

\(\operatorname{cosec}\dfrac{\pi}{3}\)\(=\)\(\dfrac{1}{\sin\dfrac{\pi}{3}}\)
\(\sin\dfrac{\pi}{3}\)\(=\)\(\dfrac{\sqrt{3}}{2}\)
\(=\)\(\dfrac{2}{\sqrt{3}}\)
\(=\)\(\dfrac{2\sqrt{3}}{3}\)

\(\operatorname{cosec}\dfrac{\pi}{3}=\dfrac{2\sqrt{3}}{3}\).

Example 3 — a second-quadrant angle
Find the exact value of \(\sec\dfrac{2\pi}{3}\).
Solution

The angle \(\dfrac{2\pi}{3}\) is in the second quadrant, where cosine is negative. Find the cosine, then take its reciprocal:

\(\cos\dfrac{2\pi}{3}\)\(=\)\(-\dfrac{1}{2}\)
\(\sec\dfrac{2\pi}{3}\)\(=\)\(\dfrac{1}{\cos\dfrac{2\pi}{3}}\)
\(=\)\(\dfrac{1}{-\dfrac{1}{2}}\)
\(=\)\(-2\)

\(\sec\dfrac{2\pi}{3}=-2\).

Unit circle and the reciprocal ratios A unit circle with a radius drawn to a point P at 60 degrees. The horizontal coordinate of P is cos and the vertical coordinate is sin, so sec is one over cos and cosec is one over sin. x y P θ cos sin
Example 4 — simplify an expression
Simplify \(\dfrac{\sec x}{\operatorname{cosec} x}\).
Solution

Write each function as a reciprocal, then divide the fractions:

\(\dfrac{\sec x}{\operatorname{cosec} x}\)\(=\)\(\dfrac{\dfrac{1}{\cos x}}{\dfrac{1}{\sin x}}\)
\(=\)\(\dfrac{1}{\cos x}\times\dfrac{\sin x}{1}\)
\(=\)\(\dfrac{\sin x}{\cos x}\)
\(=\)\(\tan x\)

\(\dfrac{\sec x}{\operatorname{cosec} x}=\tan x\).

Common pitfalls

Swapping secant and cosecant. The names do not match the ratios: secant goes with cosine and cosecant goes with sine. A quick check — co-secant, co-sine? No: cosec is \(1/\sin\).
Reading it as an inverse. \(\sec x\) is \(1/\cos x\), not \(\cos^{-1}x\) and not \(1/x\). The angle stays inside the function.
Forgetting the undefined points. Before you divide, check the base ratio is not zero: \(\sec x\) is undefined at \(x=\dfrac{\pi}{2},\dfrac{3\pi}{2}\); \(\operatorname{cosec} x\) and \(\cot x\) at \(x=0,\pi,2\pi\).
Leaving a surd on the bottom. Rationalise: \(\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}\), and never round an exact value to a decimal.

Frequently asked questions

What are sec, cosec and cot?

They are the reciprocal trigonometric functions: \(\sec x=\dfrac{1}{\cos x}\), \(\operatorname{cosec} x=\dfrac{1}{\sin x}\) and \(\cot x=\dfrac{1}{\tan x}=\dfrac{\cos x}{\sin x}\).

Is cosec the reciprocal of sine or cosine?

Cosecant is the reciprocal of sine: \(\operatorname{cosec} x=\dfrac{1}{\sin x}\). Secant is the reciprocal of cosine. The names deliberately do not line up with the ratios.

Where is sec x undefined?

Wherever \(\cos x=0\), because you cannot divide by zero. On \(0\le x\le 2\pi\) that is \(x=\dfrac{\pi}{2}\) and \(x=\dfrac{3\pi}{2}\).

How do I find sec of pi on 3?

Take the reciprocal of \(\cos\dfrac{\pi}{3}=\dfrac{1}{2}\), giving \(\sec\dfrac{\pi}{3}=\dfrac{1}{1/2}=2\).

Is sec x the same as cos inverse x?

No. \(\sec x=\dfrac{1}{\cos x}\) is a reciprocal, while \(\cos^{-1}x\) is the inverse cosine that returns an angle. They are different functions.

How do I get cot from cos and sin?

Cotangent is cosine divided by sine: \(\cot x=\dfrac{\cos x}{\sin x}\), which is also \(\dfrac{1}{\tan x}\).