Reciprocal trigonometric functions and the Pythagorean identity
Meet the reciprocal trigonometric functions — secant, cosecant and cotangent — and the Pythagorean identities for Year 11 Specialist Mathematics in Queensland (QCAA).
You will derive the reciprocal identities from sin squared plus cos squared equals one, then use them with the quadrant of an angle to find exact values, simplify expressions and prove identities — core trigonometry skills for the rest of the course.
Theory
The reciprocal trigonometric functions — secant, cosecant and cotangent — and the Pythagorean identities are core tools of Year 11 Specialist Mathematics (QCAA, Queensland). Starting from \(\sin^2 x+\cos^2 x=1\), this page derives \(1+\tan^2 x=\sec^2 x\) and \(1+\cot^2 x=\operatorname{cosec}^2 x\), then uses them to find exact values and simplify expressions.
Each of the three basic ratios has a reciprocal function. The secant is the reciprocal of cosine, \(\sec x=\dfrac{1}{\cos x}\); the cosecant is the reciprocal of sine, \(\operatorname{cosec} x=\dfrac{1}{\sin x}\); and the cotangent is the reciprocal of tangent, \(\cot x=\dfrac{1}{\tan x}=\dfrac{\cos x}{\sin x}\). (In Australian schools cosecant is written cosec.)
Notice the pairing is not the obvious one: cosec goes with sine and sec goes with cosine. Each is undefined wherever its denominator is zero — \(\sec x\) has no value when \(\cos x=0\), and \(\operatorname{cosec} x\) has no value when \(\sin x=0\).
The Pythagorean identity \(\sin^2 x+\cos^2 x=1\) holds for every angle \(x\); on the unit circle it is just Pythagoras' theorem applied to the point \((\cos x,\sin x)\). Dividing it through by \(\cos^2 x\) or by \(\sin^2 x\) produces the two reciprocal Pythagorean identities.
Dividing by \(\cos^2 x\) gives \(1+\tan^2 x=\sec^2 x\); dividing by \(\sin^2 x\) gives \(1+\cot^2 x=\operatorname{cosec}^2 x\). Combined with the quadrant of the angle (which fixes each sign), these let you find every ratio from just one given value.
The three reciprocal functions:
The Pythagorean identity and its two reciprocal forms:
Finding a ratio from one given value
- Choose the identity that links what you are given to what you want: \(1+\tan^2 x=\sec^2 x\) pairs tan with sec, and \(1+\cot^2 x=\operatorname{cosec}^2 x\) pairs cot with cosec.
- Substitute and simplify to get the square of the unknown (for example \(\tan^2 x=\sec^2 x-1\)).
- Take the square root to find the size of the answer.
- Fix the sign from the quadrant: decide whether the required function is positive or negative in that quadrant, and attach the correct sign.
Take the reciprocal of \(\cos x\) to get \(\sec x\):
| \(\sec x\) | \(=\) | \(\dfrac{1}{\cos x}\) |
| \(=\) | \(\dfrac{1}{3/5}\) | |
| \(=\) | \(\dfrac{5}{3}\) |
Now use \(1+\tan^2 x=\sec^2 x\); \(x\) is acute so \(\tan x>0\):
| \(\tan^2 x\) | \(=\) | \(\sec^2 x-1\) |
| \(=\) | \(\dfrac{25}{9}-\dfrac{9}{9}\) | |
| \(=\) | \(\dfrac{16}{9}\) | |
| \(\tan x\) | \(=\) | \(\sqrt{\dfrac{16}{9}}\) |
| \(=\) | \(\dfrac{4}{3}\) |
\(\sec x=\dfrac{5}{3}\) and \(\tan x=\dfrac{4}{3}\).
Choose the identity that pairs cot with cosec, then rearrange for \(\cot^2 x\):
| \(1+\cot^2 x\) | \(=\) | \(\operatorname{cosec}^2 x\) |
| \(\cot^2 x\) | \(=\) | \(\operatorname{cosec}^2 x-1\) |
| \(=\) | \(\dfrac{625}{49}-\dfrac{49}{49}\) | |
| \(=\) | \(\dfrac{576}{49}\) |
Take the positive root because \(x\) is acute, so \(\cot x>0\):
| \(\cot x\) | \(=\) | \(\sqrt{\dfrac{576}{49}}\) |
| \(=\) | \(\dfrac{24}{7}\) |
\(\cot x=\dfrac{24}{7}\).
Use \(\sec^2 x=1+\tan^2 x\) to find the square:
| \(\sec^2 x\) | \(=\) | \(1+\tan^2 x\) |
| \(=\) | \(1+\dfrac{9}{16}\) | |
| \(=\) | \(\dfrac{25}{16}\) |
In the fourth quadrant \(\cos x>0\), so \(\sec x>0\); take the positive root:
| \(\sec x\) | \(=\) | \(+\sqrt{\dfrac{25}{16}}\) |
| \(=\) | \(\dfrac{5}{4}\) |
\(\sec x=\dfrac{5}{4}\).
Start from the left side and write each term over a common denominator:
| \(\sec^2 x+\operatorname{cosec}^2 x\) | \(=\) | \(\dfrac{1}{\cos^2 x}+\dfrac{1}{\sin^2 x}\) |
| \(=\) | \(\dfrac{\sin^2 x+\cos^2 x}{\sin^2 x\cos^2 x}\) |
Apply \(\sin^2 x+\cos^2 x=1\) in the numerator, then split the fraction:
| \(=\) | \(\dfrac{1}{\sin^2 x\cos^2 x}\) | |
| \(=\) | \(\dfrac{1}{\cos^2 x}\times\dfrac{1}{\sin^2 x}\) | |
| \(=\) | \(\sec^2 x\,\operatorname{cosec}^2 x\) |
The left side equals the right side, so the identity holds.
Common pitfalls
Frequently asked questions
What are the reciprocal trigonometric functions?
They are secant \(\sec x=\dfrac{1}{\cos x}\), cosecant \(\operatorname{cosec} x=\dfrac{1}{\sin x}\) and cotangent \(\cot x=\dfrac{1}{\tan x}=\dfrac{\cos x}{\sin x}\).
Why does cosec go with sine and sec with cosine?
It is just the naming convention: cosecant is defined as one over sine, and secant as one over cosine. The prefixes cross over, so it is worth memorising.
How do you derive \(1+\tan^2 x=\sec^2 x\)?
Start from \(\sin^2 x+\cos^2 x=1\) and divide every term by \(\cos^2 x\). This gives \(\tan^2 x+1=\sec^2 x\).
What are the three Pythagorean identities?
They are \(\sin^2 x+\cos^2 x=1\), \(1+\tan^2 x=\sec^2 x\) and \(1+\cot^2 x=\operatorname{cosec}^2 x\).
How do I know whether the answer is positive or negative?
Take the square root for the size, then use the quadrant of the angle to choose the sign, since each function is positive in some quadrants and negative in others.
Is \(\operatorname{cosec} x\) the same as \(\sin^{-1}x\)?
No. \(\operatorname{cosec} x\) is the reciprocal \(\dfrac{1}{\sin x}\); \(\sin^{-1}x\) means the inverse-sine (arcsine) function. They are different things.