The four remaining circular functions defined from the same two coordinates: secant and cosecant as the reciprocals of cosine and sine, tangent and cotangent as their quotients, each with the restriction its denominator forces. Covers the geometric origin of the names tangent and secant, the reciprocal and quotient identities, the Generalized Reference Angle Theorem extending the reference-angle method to all six, the two further Pythagorean identities obtained by dividing the first through by a square, and the techniques for verifying an identity including Pythagorean conjugates.
Subject: Trigonometry · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Trigonometry · Chapter 10 — Foundations of Trigonometry
§10.3 The Six Circular Functions and Fundamental Identities, pp. 744-752
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 744-752 — the pages these objectives are drawn from
Warm-up
You can find the cosine and sine of any special angle. Four more functions are about to be defined, and every one of them is built from those two.
Discussion prompt
At the angle pi over two the point on the unit circle is zero comma one. If someone defined a new function as one divided by the x-coordinate, what would it be at pi over two, and what does that tell you about the new function?
Hint: Substitute the x-coordinate and look at what you are being asked to divide by.
Answer:
\[ x = 0 \quad\Longrightarrow\quad \frac{1}{x} = \frac{1}{0} \]
It is undefined. And that is not a defect of the example — it is the defining feature of the four functions in this lesson. Cosine and sine are defined for every angle without exception; the other four each carry a denominator, and each is undefined wherever that denominator vanishes.
So a genuinely new question arrives with this lesson, one that never came up before: for which angles does this even exist?
Concept
The point on the unit circle has two coordinates, and those two coordinates are already the cosine and the sine. The remaining four circular functions are simply the reciprocals and the quotients of them, so nothing new is being measured.
circular functions — The six functions cosine, sine, secant, cosecant, tangent and cotangent, each defined from the coordinates x and y of the point where an angle's terminal side meets the unit circle.
\[ \sec(\theta) = \frac{1}{x}, \quad \csc(\theta) = \frac{1}{y}, \quad \tan(\theta) = \frac{y}{x}, \quad \cot(\theta) = \frac{x}{y} \]
What is new is not information but convenience, and one genuine complication: four of the six now have angles where they do not exist, so every statement about them has to carry a condition.
Figure (svg): A table of the six circular functions, each defined from the coordinates x and y of the point on the unit circle, with the excluded case noted for the four that have one
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 744-744
Section
Section 1
Concept
Each of the four new functions is a fraction, and a fraction needs a nonzero denominator. Reading the definitions carefully tells you exactly which angles each function excludes.
A common confusion: secant pairs with cosine, not with sine, despite the mismatched initials. The pairing is by which coordinate is downstairs, and it is worth checking rather than guessing.
Figure (svg): A table of the six circular functions, each defined from the coordinates x and y of the point on the unit circle, with the excluded case noted for the four that have one
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 744-745
Picture it
Tangent and secant are not arbitrary labels. For an acute angle they are lengths in this picture.
Figure (svg): The unit circle with the vertical tangent line at x equals one, showing that the tangent of an acute angle is the height at which the terminal side crosses that line, and the secant is the length of the terminal side out to that crossing
Tangent comes from the Latin for touch, and the line x equals one touches the circle at exactly one point. Secant comes from the word for cut, and the terminal side's line cuts the circle in two places.
Worked example
Example 10.3.1, parts 1 and 2. Find the cosine or sine first, then reciprocate.
\[ \text{Find } \sec(60^\circ) \text{ and } \csc\left(\frac{7\pi}{4}\right). \]
Find the cosine of 60 degrees
Why: This is one of the three special values from Lesson 10.2a.
\[ \cos 60 ^\circ = \frac{1}{2} \]
Reciprocate it
Why: Secant is one over the cosine.
\[ \sec 60 ^\circ = \frac{1}{\frac{1}{2}} = 2 \]
Find the sine of seven pi over four
Why: Quadrant four, reference angle pi over four, so the sine is negative root two over two.
\[ \sin = -\sqrt{2} / 2 \]
Reciprocate and rationalise
Why: One over negative root two over two is negative two over root two, which simplifies to negative root two.
\[ \csc = -\sqrt{2} \]
Figure (svg): The solution to Worked example evaluate a secant and a cosecant shown as a ladder of expressions, one row per legal move
\[ \sec(60^\circ) = 2 \qquad \csc\left(\frac{7\pi}{4}\right) = -\sqrt{2} \]
Verify: check the sizes
Why: A cosine of one half is small, so its reciprocal should be large, and 2 is larger than 1. In fact a secant or cosecant can never lie strictly between negative one and one, since it is the reciprocal of something with absolute value at most 1. Negative root two is about negative 1.414, comfortably outside that band.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 746-746
Sorting
Ask which coordinate is downstairs, then ask where that coordinate vanishes.
Sort into buckets
Sort each function by where it fails to exist.
Worked example
Example 10.3.1, part 4. The answer here is not a number.
\[ \text{Find } \tan(\theta) \text{ if } \theta \text{ is coterminal with } \frac{3\pi}{2}. \]
Find the cosine and sine
Why: Coterminal angles share the point, and at three pi over two the point is zero comma negative one.
\[ \cos = 0, \sin = -1 \]
Write the quotient
Why: Tangent is the sine over the cosine, which here is negative one over zero.
\[ \tan = -\frac{1}{0} \]
Say what that means
Why: Division by zero is undefined, so the tangent does not exist at this angle.
Note the pattern
Why: Tangent is undefined exactly where the cosine is zero, which is exactly the quadrantal angles on the y-axis.
Figure (svg): The solution to Worked example recognise an undefined value shown as a ladder of expressions, one row per legal move
\[ \tan(\theta) \text{ is undefined, since } \cos(\theta) = 0 \]
Verify: check the companion function
Why: Secant shares the denominator, so it must also be undefined here — and indeed one over zero fails the same way. Meanwhile cosecant is one over negative one, which is negative one, and cotangent is zero over negative one, which is zero. Both of those exist, because their denominator is the sine, which is not zero here.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 746-746
Trap
\[ \sec(\theta) = \frac{1}{\sin(\theta)} \]
Pair the functions by their initial letters, s with s
Why: Secant and sine both begin with s, and cosecant and cosine both begin with c, so the pairing looks obvious.
It is exactly backwards. This would make the secant of pi over two equal to one, when in fact it is undefined there.
\[ \sec(\theta) = \frac{1}{\cos(\theta)}, \qquad \csc(\theta) = \frac{1}{\sin(\theta)} \]
Pair by which coordinate sits downstairs
Why: Secant is one over x, and x is the cosine. Cosecant is one over y, and y is the sine.
A memory hook that works: look at the third letter. The third letter of secant is c, and it pairs with cosine; the third letter of cosecant is s, and it pairs with sine. Or simply remember that the pairing is deliberately crossed, because a name beginning with co pairs with a name not beginning with co.
Prediction
An angle has cosine equal to 0.2.
Predict first
What can you say about its secant?
Correct: It is 5, a value larger than 1.
Why: Secant is the reciprocal of cosine, so it is one over 0.2, which is 5. The general principle is worth carrying: because a cosine always lies between negative one and one, its reciprocal always lies outside that band. A small cosine gives a large secant, and as the cosine approaches zero the secant grows without bound — which is precisely the behaviour that produces the vertical asymptotes in the graph of Lesson 10.5b.
Matching
Three reciprocal pairs, and the pairing is crossed on purpose.
Match the pairs
Why: Cosine is x and secant is one over x; sine is y and cosecant is one over y; tangent is y over x and cotangent is x over y, which is its reciprocal. Notice that only the third pair has matching prefixes, and the first two are deliberately crossed — a name beginning with co pairs with a name that does not.
Socratic
The picture showed that for an acute angle, tangent is the height at which the terminal side meets the vertical line x equals one.
Discussion prompt
Reconstruct the similar-triangles argument that proves this. Why does the answer come out as y over x?
Hint: Compare the triangle with hypotenuse to the unit circle against the triangle with hypotenuse out to the tangent line.
Answer:
Drop perpendiculars from P, the point on the unit circle, and from Q, the point on the line x equals one. The two right triangles share the angle at the origin, so they are similar.
\[ \frac{y'}{y} = \frac{1}{x} \quad\Longrightarrow\quad y' = \frac{y}{x} = \tan(\theta) \]
The horizontal legs are x and 1 respectively, so the ratio of similarity is one over x, and the vertical legs must be in that same ratio. Hence the height on the tangent line is y over x, which is the definition of the tangent.
The same argument on the hypotenuses gives the secant as one over x, which is why the two functions share a denominator and share their undefined angles.
Section
Section 2
Concept
The definitions can be rewritten with cosine and sine in place of x and y, which turns them into identities. The practical consequence is that any expression in the six functions can be rewritten in just two of them.
Theorem 10.6 — The reciprocal and quotient identities: each of the four remaining circular functions expressed in terms of cosine and sine, valid wherever the denominator involved is nonzero.
\[ \sec\theta = \frac{1}{\cos\theta}, \; \csc\theta = \frac{1}{\sin\theta}, \; \tan\theta = \frac{\sin\theta}{\cos\theta}, \; \cot\theta = \frac{\cos\theta}{\sin\theta} \]
Converting everything to cosine and sine is the single most reliable strategy in this chapter. It is not always the shortest route, but it is the one that always works when nothing else is obvious.
| Angle | tangent | cotangent |
|---|---|---|
| 0 | 0 | undefined |
| pi over 6 | root 3 over 3 | root 3 |
| pi over 4 | 1 | 1 |
| pi over 3 | root 3 | root 3 over 3 |
| pi over 2 | undefined | 0 |
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 745-747
Picture it
Six functions, but only three patterns, because a reciprocal always carries the sign of what it reciprocates.
Figure (svg): A chart showing the sign of all six circular functions in each of the four quadrants
Tangent and cotangent are positive where x and y have the same sign, which is quadrants one and three. That is a genuinely different pattern from either cosine or sine, and it is what makes tangent equations have solutions half a revolution apart.
Worked example
Example 10.3.1, part 5. Reciprocate, then use the Pythagorean identity, then use the quadrant.
\[ \text{If } \csc(\theta) = -\sqrt{5} \text{ and } \theta \text{ is in Quadrant IV, find } \cos(\theta). \]
Convert the cosecant to a sine
Why: Cosecant is one over sine, so sine is one over cosecant.
\[ \sin \theta = -1 / \sqrt{5} \]
Rationalise
Why: Multiply top and bottom by root five.
\[ \sin \theta = -\sqrt{5} / 5 \]
Substitute into the Pythagorean identity
Why: One fifth subtracted from 1 leaves four fifths.
\[ \cos ^{2} \theta = \frac{4}{5} \]
Take the root and choose the sign
Why: The root of four fifths is two over root five; quadrant four has positive cosine.
\[ \cos \theta = 2 \sqrt{5} / 5 \]
Figure (svg): The solution to Worked example from a cosecant to a cosine shown as a ladder of expressions, one row per legal move
\[ \cos(\theta) = \frac{2\sqrt{5}}{5} \]
Verify: check the identity and the quadrant
Why: Four fifths plus one fifth is 1. And quadrant four wants a positive cosine with a negative sine, which is exactly what came out. Note also that the cosecant being negative already told us the sine was negative, consistent with quadrant four.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 746-746
Faded example
Express the cotangent of theta in terms of cosine and sine, then evaluate it at pi over three.
Fill in the blanks
\cot(\theta) = \fraccos theta}sqrt3/2 \quad\Longrightarrow\quad \cot\left(\fracsqrt3/3___\right) = \frac______} = ___
Why: Cotangent is x over y, which is the cosine over the sine. At pi over three the cosine is one half and the sine is root three over two, so the quotient is one over root three, which rationalises to root three over three. Notice this is the reciprocal of the tangent of pi over three, which is root three, exactly as the reciprocal identity requires.
Worked example
Example 10.3.1, part 6. The book flags the tempting wrong move here, and it is worth seeing.
\[ \text{If } \tan(\theta) = 3 \text{ and } \pi < \theta < \frac{3\pi}{2}, \text{ find } \sin(\theta). \]
Write the tangent as a quotient
Why: This does NOT mean the sine is 3 and the cosine is 1 — that pair fails the Pythagorean identity.
\[ \sin / \cos = 3 \]
Express one function in terms of the other
Why: Rearranged, the cosine is one third of the sine.
\[ \cos \theta = (\frac{1}{3}) \sin \theta \]
Substitute into the Pythagorean identity
Why: Sine squared plus one ninth of sine squared equals 1.
\[ (\frac{10}{9}) \sin ^{2} \theta = 1 \]
Solve and choose the sign from the quadrant
Why: Sine squared is nine tenths; the interval names quadrant three, where the sine is negative.
\[ \sin \theta = -3 \sqrt{10} / 10 \]
Figure (svg): The solution to Worked example from a tangent to a sine shown as a ladder of expressions, one row per legal move
\[ \sin(\theta) = -\frac{3\sqrt{10}}{10} \]
Verify: reconstruct the tangent
Why: The cosine is one third of that, namely negative root ten over ten. Dividing the sine by the cosine gives negative three root ten over ten divided by negative root ten over ten, which is 3 — the tangent we were given. And both values are negative, matching quadrant three.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 746-746
Error analysis
A student is told the tangent of theta is 3 and writes down the cosine and sine directly.
Annotate
On: \( \tan(\theta) = 3 = \frac{3}{1} \quad\Longrightarrow\quad \sin(\theta) = 3, \; \cos(\theta) = 1 \)
The values three and one do describe a legitimate point on the terminal side, just not one on the unit circle. Dividing both by the radius, root ten, gives the actual values — which is exactly Theorem 10.3 from the previous lesson, and is a perfectly good alternative route to the answer.
Sorting
Tangent is y over x, so it is positive when the two coordinates agree in sign.
Sort into buckets
Sort each angle by the sign of its tangent.
Estimation
An angle is just slightly less than pi over two, say 1.5 radians.
Predict first
Roughly how big is its tangent?
Correct: About 14.
Why: At 1.5 radians the sine is about 0.997 and the cosine is about 0.0707, so the quotient is about 14.1. As the angle approaches pi over two the cosine approaches zero while the sine approaches one, so the tangent grows without bound. The distractor 0.07 is the cosine itself, and it is the reciprocal relationship that makes the tangent large rather than small — a good reminder that a tiny denominator means a huge quotient.
Counterexample
A student proposes: since tangent is sine over cosine, the tangent is undefined exactly where the sine is undefined.
Discussion prompt
Explain why this is doubly wrong, and state correctly where the tangent is undefined.
Hint: Ask where the sine is undefined at all.
Answer:
The first error is that the sine is never undefined. It is a coordinate of a point that exists for every angle, so the claim's condition never occurs and would predict that tangent is always defined.
The second is that the sine is in the numerator of the tangent, and a numerator cannot make a fraction undefined. Only the denominator can, and the denominator is the cosine.
\[ \tan(\theta) \text{ undefined} \iff \cos(\theta) = 0 \iff \theta = \tfrac{\pi}{2} + \pi k \]
A zero numerator is perfectly fine and simply makes the tangent zero, which is what happens at zero and at pi.
Section
Section 3
Concept
An equation in secant, cosecant, tangent or cotangent can always be converted into one in cosine or sine. Alternatively the Generalized Reference Angle Theorem handles it directly, using the sign of the value to pick the quadrants.
Theorem 10.7 — The Generalized Reference Angle Theorem: the value of any circular function at an angle equals, up to sign, its value at the reference angle, with the sign determined by the quadrant.
That combination is a genuine simplification and not just a cosmetic one: it reflects the fact that tangent repeats every half revolution, which is why its graph in Lesson 10.5c has period pi rather than two pi.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 747-748
Picture it
Solving needs the sign chart read backwards: given the sign of a value, which quadrants can the angle be in?
Figure (svg): A chart showing the sign of all six circular functions in each of the four quadrants
Cosine positive means quadrants one and four, which are adjacent. Tangent positive means quadrants one and three, which are opposite. That difference is the whole reason tangent families combine and cosine families do not.
Worked example
Example 10.3.2, part 1. Reciprocate and the problem becomes one you have already solved.
\[ \text{Solve } \sec(\theta) = 2 \text{ for all } \theta. \]
Convert to a cosine equation
Why: Secant is one over cosine, so one over cosine equals 2.
\[ 1 / \cos \theta = 2 \]
Solve for the cosine
Why: Reciprocating both sides.
\[ \cos \theta = \frac{1}{2} \]
Recognise the equation
Why: This is exactly the equation solved in Example 10.2.5 part 1.
Quote the two families
Why: Positive cosine gives quadrants one and four, with reference angle pi over three.
\[ \frac{\pi}{3} + 2 \pi k, 5 \pi / 3 + 2 \pi k \]
Figure (svg): The solution to Worked example solve a secant equation shown as a ladder of expressions, one row per legal move
\[ \theta = \frac{\pi}{3} + 2\pi k \quad\text{or}\quad \theta = \frac{5\pi}{3} + 2\pi k, \quad k \in \mathbb{Z} \]
Verify: check a member of each family
Why: At pi over three the cosine is one half so the secant is 2. At five pi over three the cosine is also one half, so the secant is again 2. Both families check out.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 747-747
Discrimination
Combination works exactly when the two base solutions differ by pi.
Sort into buckets
Sort each equation by whether its two solution families combine into one stepping by pi.
Worked example
Example 10.3.2, part 3. The combination at the end is the interesting part.
\[ \text{Solve } \cot(\theta) = -1 \text{ for all } \theta. \]
Find the reference angle from the size
Why: The cotangent of pi over four is 1, so the reference angle is pi over four.
\[ \alpha = \frac{\pi}{4} \]
Find the quadrants from the sign
Why: Cotangent is x over y, negative when the coordinates have opposite signs, which is quadrants two and four.
Write one solution per quadrant
Why: Quadrant two gives three pi over four; quadrant four gives seven pi over four.
\[ 3 \pi / 4\text{ and } 7 \pi / 4 \]
Combine the two families
Why: The two solutions differ by exactly pi, so a single family stepping by pi captures both.
\[ \theta = 3 \pi / 4 + \pi k \]
Figure (svg): The solution to Worked example solve a cotangent equation and combine the families shown as a ladder of expressions, one row per legal move
\[ \theta = \frac{3\pi}{4} + \pi k, \quad k \in \mathbb{Z} \]
Verify: expand the combined family
Why: Taking k equal to 0 gives three pi over four; k equal to 1 gives seven pi over four; k equal to 2 gives eleven pi over four, which is coterminal with three pi over four. So the single family produces exactly the angles the two separate families did, alternating between them.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 748-748
Trap
\[ \cos(\theta) = \tfrac{1}{2} \quad\Longrightarrow\quad \theta = \tfrac{\pi}{3} + \pi k \]
Combine the two families into one stepping by pi, as was done for cotangent
Why: The combination worked there, so it looks like a general simplification.
Taking k equal to 1 gives pi over three plus pi, which is four pi over three. The cosine there is negative one half, not one half. The family contains angles that do not solve the equation.
\[ \theta = \tfrac{\pi}{3} + 2\pi k \quad\text{or}\quad \theta = \tfrac{5\pi}{3} + 2\pi k \]
Keep the two families separate for cosine and sine
Why: The two solution quadrants for a cosine equation are adjacent, not opposite, so the solutions are not half a revolution apart.
The test is simple: combine only when the two base solutions differ by exactly pi. For cotangent equal to negative one they were three pi over four and seven pi over four, a difference of pi, so the combination is valid. Here they are pi over three and five pi over three, differing by four pi over three, and no single family with a constant step captures both.
Fill the middle
Solve the equation tangent of theta equals root three.
Fill in the blanks
\alpha = pi/3, \quad \textIpi \text___ \;\Longrightarrow\; \theta = \frac______ + ___ k
Why: The tangent of pi over three is root three, so that is the reference angle. A positive tangent means the coordinates share a sign, which happens in quadrants one and three. The two base solutions are pi over three and four pi over three, differing by exactly pi, so they combine into a single family stepping by pi.
Elimination
The equation is cosecant of theta equals 2.
Eliminate the wrong options
Which describes all the solutions?
Survives elimination: B
Why: Cosecant equal to 2 means the sine is one half, a positive value, so the horizontal line sits above the axis and cuts in quadrants one and two. The reference angle for a sine of one half is pi over six, giving base solutions pi over six and five pi over six. These are not half a revolution apart, so the families stay separate and each carries plus two pi k.
Missing information
A problem says: solve secant of theta equals one half.
Discussion prompt
Work it through and say what happens. What does the answer tell you about the range of the secant function?
Hint: Convert to a cosine equation and check the value against the range of cosine.
Answer:
\[ \sec(\theta) = \tfrac{1}{2} \;\Longrightarrow\; \frac{1}{\cos(\theta)} = \tfrac{1}{2} \;\Longrightarrow\; \cos(\theta) = 2 \]
No angle has a cosine of 2, so the equation has no solutions at all. Nothing was missing from the problem; the answer is simply empty.
The general fact this exposes: since a cosine always lies between negative one and one, its reciprocal always has absolute value at least one. So the range of secant is everything outside the open interval from negative one to one, and the same holds for cosecant. An equation setting either of them to a value strictly between negative one and one is unsolvable, and recognising that instantly saves a page of work.
Section
Section 4
Concept
The second and third Pythagorean identities are not independent facts to memorise. Each is the first one divided through by a square, and doing the division once makes both permanent.
\[ \cos^2\theta + \sin^2\theta = 1 \]
The restrictions are inherited from the division, not added by hand: you may not divide by zero, so the second identity says nothing at pi over two and the third says nothing at zero.
Figure (svg): The three Pythagorean identities stacked as cards, the first for cosine and sine, the second for tangent and secant, and the third for cotangent and cosecant, with the restriction on each
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 748-749
Picture it
Read them as one identity and two consequences, not as three things to memorise.
Figure (svg): The three Pythagorean identities stacked as cards, the first for cosine and sine, the second for tangent and secant, and the third for cotangent and cosecant, with the restriction on each
If you ever forget the second or third, write the first and divide. It takes ten seconds and it cannot produce a wrong sign, which recall from memory sometimes can.
Worked example
Do this once and you will not need to remember the result.
\[ \text{Derive } 1 + \tan^2(\theta) = \sec^2(\theta) \text{ from the Pythagorean identity.} \]
Start from the first identity
Why: The one that holds for every angle.
\[ \cos ^{2} + \sin ^{2} = 1 \]
Divide every term by cosine squared
Why: Legal wherever the cosine is nonzero, which is where the result will hold.
\[ 1 + \sin ^{2} / \cos ^{2} = 1 / \cos ^{2} \]
Recognise the middle term
Why: Sine over cosine is the tangent, so its square is tangent squared.
\[ \sin ^{2} / \cos ^{2} = \tan ^{2} \]
Recognise the right-hand term
Why: One over cosine is the secant, so one over cosine squared is secant squared.
\[ 1 / \cos ^{2} = \sec ^{2} \]
Figure (svg): The solution to Worked example derive the second identity shown as a ladder of expressions, one row per legal move
\[ 1 + \tan^2(\theta) = \sec^2(\theta), \quad \cos(\theta) \ne 0 \]
Verify: test it at pi over four
Why: There the tangent is 1 and the secant is root two. The left side is 1 plus 1, which is 2; the right side is root two squared, which is 2. They agree.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 748-749
Matching
Every one of these is an alternate form of one of the three identities.
Match the pairs
Why: Each is one of the three identities with a term moved across. The first and fourth come from rearranging the original; the second comes from the tangent-secant identity written as a difference; the third comes from the cotangent-cosecant identity. Spotting these rearrangements inside a larger expression is the main skill in identity verification, and it is worth being able to read them in either direction.
Worked example
Recognising an alternate form is what makes these useful.
\[ \text{Simplify } \frac{\sec^2(\theta) - 1}{\tan(\theta)}. \]
Recognise the numerator
Why: Rearranging the second identity gives secant squared minus one equals tangent squared.
\[ \sec ^{2} - 1 = \tan ^{2} \]
Substitute
Why: The numerator becomes a single term.
\[ \tan ^{2} / \tan \]
Cancel
Why: Valid wherever the tangent is nonzero, which the expression already required.
State the restriction
Why: The original expression needed the tangent defined and nonzero.
\[ \cos\text{ not } 0\text{ and } \sin\text{ not } 0 \]
Figure (svg): The solution to Worked example use an identity to simplify shown as a ladder of expressions, one row per legal move
\[ \frac{\sec^2(\theta) - 1}{\tan(\theta)} = \frac{\tan^2(\theta)}{\tan(\theta)} = \tan(\theta) \]
Verify: test at pi over three
Why: The secant there is 2 and the tangent is root three. The original is 4 minus 1 over root three, which is 3 over root three, which is root three. And the simplified form is the tangent, root three. They agree.
Error analysis
A student tries to derive the third Pythagorean identity.
Annotate
On: \( \frac{\cos^2\theta}{\cos^2\theta} + \frac{\sin^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta} \;\Longrightarrow\; 1 + \cot^2\theta = \csc^2\theta \)
A quick check on any claimed identity: substitute a convenient angle. At pi over four the cotangent is 1 and the cosecant is root two, so the third identity reads 1 plus 1 equals 2, which is fine — but the derivation above did not produce it, and checking the derivation is a different task from checking the result.
Faded example
Derive the third Pythagorean identity.
Fill in the blanks
\fracsin^2 thetacot^2 theta} + \fraccsc^2 theta___} = \frac______} \;\Longrightarrow\; ___ + 1 = ___
Why: Dividing through by sine squared makes the first term cosine squared over sine squared, which is cotangent squared, the second term equal to 1, and the right-hand side one over sine squared, which is cosecant squared. The restriction that the sine be nonzero is inherited from the division and is not an extra hypothesis.
Two truths and a lie
Rule out the statements that are true. The survivor is the false one.
Eliminate the wrong options
One of these statements about the Pythagorean identities is wrong.
Survives elimination: B
Why: Statement B is false. Only the first identity holds for every angle, because it involves only the two coordinates and those always exist. The second and third involve functions with denominators and are silent wherever those denominators vanish. Stating an identity without its restriction is a real error rather than a pedantic one: it invites you to apply the identity at an angle where one of its terms does not exist.
Edge cases
The second identity says one plus tangent squared equals secant squared.
Discussion prompt
Both sides grow without bound as the angle approaches pi over two. What does the identity tell you about how their growth compares, and why can the secant never be smaller than the tangent in absolute value?
Hint: Rearrange it as a difference.
Answer:
\[ \sec^2(\theta) - \tan^2(\theta) = 1 \]
Their squares differ by exactly 1, at every angle, no matter how large both become. So as the angle approaches pi over two and both blow up, the gap between their squares stays fixed — meaning the two functions become proportionally closer and closer, even as each grows without bound.
Secant squared is always tangent squared plus one, so secant squared is always strictly larger, and taking roots, the secant always exceeds the tangent in absolute value. Geometrically this is the statement that the hypotenuse of a right triangle is longer than either leg — which is what the picture of the tangent line was showing all along.
Section
Section 5
Concept
To verify an identity you start with one side, usually the more complicated one, and transform it by legal steps until it becomes the other. You do not treat it as an equation and operate on both sides, because that assumes what you are trying to prove.
The book makes a practical point worth repeating: even in an exam, working on the side with more structure earns more partial credit, because there are more recognisable steps to take.
Figure (svg): Two columns pairing each Pythagorean conjugate with the single term its product simplifies to
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 749-751
Picture it
These pairs are worth recognising on sight. Multiplying by the partner is the move that unsticks most stuck verifications.
Figure (svg): Two columns pairing each Pythagorean conjugate with the single term its product simplifies to
It is exactly the same technique as rationalising a denominator containing a surd, or multiplying by a complex conjugate. Only the identity doing the collapsing is different.
Worked example
Example 10.3.3, part 6. The mismatch between one minus cosine and one plus cosine is the hint.
\[ \text{Verify } \frac{\sin(\theta)}{1 - \cos(\theta)} = \frac{1 + \cos(\theta)}{\sin(\theta)}. \]
Notice the conjugate pair
Why: The left denominator is one minus cosine and the right numerator is one plus cosine, which suggests multiplying by that partner.
Multiply top and bottom of the left side by one plus cosine
Why: Multiplying by a form of 1 changes nothing but the appearance.
\[ \times(1 + \cos) / (1 + \cos) \]
Expand the denominator as a difference of squares
Why: One minus cosine times one plus cosine is one minus cosine squared.
\[ \text{denominator } = 1 - \cos ^{2} \]
Apply the Pythagorean identity and cancel
Why: One minus cosine squared is sine squared, and one factor of sine cancels with the numerator.
\[ = (1 + \cos) / \sin \]
Figure (svg): The solution to Worked example a verification using a conjugate shown as a ladder of expressions, one row per legal move
\[ \frac{\sin\theta}{1 - \cos\theta} \cdot \frac{1 + \cos\theta}{1 + \cos\theta} = \frac{\sin\theta(1 + \cos\theta)}{\sin^2\theta} = \frac{1 + \cos\theta}{\sin\theta} \]
Verify: test at pi over three
Why: The left side is root three over two divided by one minus one half, which is root three over two divided by one half, giving root three. The right side is one plus one half divided by root three over two, which is three halves times two over root three, giving three over root three, which is root three. They agree.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 751-751
Ranking
Verify that the tangent of theta equals the sine of theta times the secant of theta, starting from the right.
Put in order
Why: This is Example 10.3.3 part 2. Beginning with the right-hand side gives more to work with, since it contains a function that can be rewritten. Replacing the secant by its reciprocal form turns the product into a single fraction, and that fraction is the quotient identity for tangent read backwards. Four short steps, each individually justified.
Worked example
Example 10.3.3, part 5. The right side is the promising one here.
\[ \text{Verify } 6\sec(\theta)\tan(\theta) = \frac{3}{1 - \sin(\theta)} - \frac{3}{1 + \sin(\theta)}. \]
Start on the right and take a common denominator
Why: The product of the two denominators is a difference of squares.
\[ \text{common denom } = 1 - \sin ^{2} \]
Combine the numerators
Why: Three plus three sine minus three plus three sine leaves six sine.
\[ \text{numerator } = 6 \sin \]
Apply the Pythagorean identity to the denominator
Why: One minus sine squared is cosine squared.
\[ = 6 \sin / \cos ^{2} \]
Split into the target form
Why: Six sine over cosine squared is six times one over cosine times sine over cosine.
\[ = 6 \sec \tan \]
Figure (svg): The solution to Worked example a verification by common denominators shown as a ladder of expressions, one row per legal move
\[ \frac{3}{1 - \sin\theta} - \frac{3}{1 + \sin\theta} = \frac{6\sin\theta}{\cos^2\theta} = 6\left(\frac{1}{\cos\theta}\right)\left(\frac{\sin\theta}{\cos\theta}\right) = 6\sec\theta\tan\theta \]
Verify: test at pi over six
Why: The right side is 3 over one half minus 3 over three halves, which is 6 minus 2, giving 4. The left side is 6 times the secant, which is 2 root three over three, times the tangent, which is root three over three, giving 6 times 2 times 3 over 9, which is 4. They agree.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 750-751
Trap
\[ \text{To verify } \frac{\sin\theta}{1 - \cos\theta} = \frac{1 + \cos\theta}{\sin\theta} \]
Cross-multiply both sides
Why: This is the standard move for solving an equation, and the statement looks like an equation.
\[ \sin^2\theta = (1 - \cos\theta)(1 + \cos\theta) \]
The step is only valid if the two sides really are equal — which is precisely what is being proved. The argument assumes its own conclusion.
\[ \frac{\sin\theta}{1 - \cos\theta} = \frac{\sin\theta}{1 - \cos\theta} \cdot \frac{1 + \cos\theta}{1 + \cos\theta} = \cdots = \frac{1 + \cos\theta}{\sin\theta} \]
Transform one side into the other by a chain of equalities
Why: Each step is justified on its own, so the chain proves the endpoints are equal rather than assuming it.
The cross-multiplied statement is not useless — it is a good way to discover the proof, and the work above shows exactly the same algebra. But a verification must be written as a one-sided chain, or it proves nothing.
Reverse engineer
A verification contains this line, applied to the expression one over one plus sine.
Fill in the blanks
\frac1 - sin thetacos^2 theta \cdot \frac___}___} = \frac______ = \frac______}
Why: The Pythagorean conjugate of one plus sine is one minus sine, and multiplying top and bottom by it turns the denominator into a difference of squares. One minus sine squared is cosine squared by the first identity, so the denominator becomes a single squared term that can then be split into secants. This is the same move as the worked example, applied to the sine version of the conjugate pair.
Prediction
You need to simplify the expression one minus cosine squared, all over sine.
Predict first
What does it simplify to?
Correct: sine.
Why: One minus cosine squared is sine squared by the first Pythagorean identity, so the expression is sine squared over sine, which is sine wherever the sine is nonzero. The whole simplification is one identity plus one cancellation. Recognising one minus cosine squared as sine squared on sight, rather than after a moment's thought, is what makes longer verifications tractable.
Explain it to yourself
The book advises starting with the more complicated side.
Discussion prompt
Explain why that is good advice rather than an arbitrary convention, and describe what you would do if both sides looked equally complicated.
Hint: Think about what a step in a verification actually does.
Answer:
Almost every legal step simplifies or rewrites something: applying an identity replaces a compound expression by a simpler one, cancelling removes a factor, combining fractions merges two terms into one. Those moves have more to grip on when there is more structure present, so a complicated side offers more available next steps and a simple side offers almost none.
If both sides look equally complicated, two things work. Transform each side separately to a common third form, which is legitimate because it is still a chain of equalities on each side independently. Or convert everything to cosine and sine on both sides, which almost always reduces the problem to ordinary fraction algebra with no trigonometry left in it.
Comparison
Fill the blanks from memory. Every row is determined by the two coordinates.
Comparison matrix
| Function | In terms of x and y | In terms of cos and sin | Undefined when |
|---|---|---|---|
| cosine | x | itself | never |
| sine | y | itself | never |
| secant | 1 over x | 1 over cosine | cosine is zero |
| cosecant | 1 over y | 1 over sine | sine is zero |
| tangent | y over x | sine over cosine | cosine is zero |
| cotangent | x over y | cosine over sine | sine is zero |
Read the last column: the four exclusions come in two pairs, because the four functions share two denominators between them.
Pattern
Whether you are evaluating, solving or verifying, the same five moves cover it.
The single most common error in this section is not arithmetic but existence: producing a number where the correct answer is undefined. Checking the denominator first costs nothing.
OpenStax Algebra and Trigonometry 2e, §7.4 The Other Trigonometric Functions §7.4
Check
An evaluation. Find the underlying cosine or sine first.
Check your understanding
What is the secant of five pi over four?
Answer: C
Why: Five pi over four is in quadrant three with reference angle pi over four, so its cosine is negative root two over two. The secant is the reciprocal, which is negative two over root two, simplifying to negative root two, about negative 1.414.
Check
An identity. Recognise the alternate form.
Check your understanding
Which single term equals cosecant squared of theta minus cotangent squared of theta?
Answer: A
Why: The third Pythagorean identity says one plus cotangent squared equals cosecant squared. Rearranging, cosecant squared minus cotangent squared equals 1. Checking at pi over four, where the cosecant is root two and the cotangent is 1, gives 2 minus 1, which is 1.
Check
Solving. Remember which families combine.
Check your understanding
Which describes all solutions of tangent of theta equals 1?
Answer: B
Why: The tangent of pi over four is 1, so that is the reference angle. A positive tangent means the coordinates share a sign, giving quadrants one and three, whose base solutions are pi over four and five pi over four. Those differ by exactly pi, so the two families combine into a single one stepping by pi.
Real world
A road sign gives a hill's gradient as 12 percent, meaning it rises 12 metres for every 100 metres travelled horizontally. A surveyor needs the angle of the slope and the actual distance travelled along the road surface per 100 metres of horizontal run.
Discussion prompt
Which of the six functions does the gradient directly report, and how do you get from it to the distance along the surface? Work both out.
Hint: Rise over run is opposite over adjacent.
Answer:
\[ \tan(\theta) = \frac{12}{100} = 0.12 \;\Longrightarrow\; \theta \approx 6.84^\circ \]
\[ \sec(\theta) = \sqrt{1 + \tan^2(\theta)} = \sqrt{1.0144} \approx 1.0072 \]
A gradient is exactly a tangent: rise over run is opposite over adjacent. Finding the angle from it needs the inverse tangent of Lesson 10.6a, but the second question does not need the angle at all.
The distance along the surface per unit of horizontal run is the secant, since secant is hypotenuse over adjacent, and the second Pythagorean identity computes it straight from the tangent with no angle in between. So 100 metres of horizontal run is about 100.72 metres of road — under one percent longer, which is why gradients this steep barely affect distance while mattering enormously for effort.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Can the secant of an angle equal one half?
Correct: No, because the secant's absolute value is always at least 1.
\[ |\cos(\theta)| \le 1 \quad\Longrightarrow\quad |\sec(\theta)| = \frac{1}{|\cos(\theta)|} \ge 1 \]
Why: The secant is the reciprocal of the cosine, and the cosine always has absolute value at most 1. Reciprocating reverses that: the secant always has absolute value at least 1, so no value strictly between negative one and one can occur. The first option states the condition correctly but a cosine of 2 is impossible, so it does not describe a real case. The fourth is simply false — the secant is negative throughout quadrants two and three.
Explain it
They have just met all six functions at once and are trying to memorise six separate definitions.
Discussion prompt
In no more than five sentences, show them how to reconstruct all six from one picture, and give them the one question that tells them instantly whether a value will be undefined.
Hint: How many independent quantities are actually on the picture?
Answer:
A usable answer: draw the point where the arm crosses the unit circle and label its two coordinates. Those two numbers are the cosine and the sine, and every other function is built from them — secant is one over the across-coordinate, cosecant is one over the up-coordinate, tangent is up over across, and cotangent is across over up. So there is really only one picture and two numbers to remember, not six definitions.
The question for undefinedness: is the denominator zero? Secant and tangent both divide by the across-coordinate, so both die on the vertical axis; cosecant and cotangent both divide by the up-coordinate, so both die on the horizontal axis. Cosine and sine never divide by anything, so they never die.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The reciprocal pairing is fixed by looking at the third letter of the name, or simply by remembering that the pairing is crossed. Undefined values are fixed by checking the denominator before computing anything. The derivations are fixed by dividing the first identity through by a square rather than trying to recall two more results. Getting stuck on a verification is fixed by having a fallback: convert everything to cosine and sine, and the problem becomes ordinary fraction algebra. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
Draw a unit circle with a point marked in the first quadrant and its two coordinates labelled x and y. Around it, write all six function definitions in terms of x and y, and beside each of the four that need one, write its restriction. Underneath, rewrite the same six in terms of cosine and sine. In the top right corner, write the first Pythagorean identity and show, with two arrows and two short divisions, how it produces the other two, noting the restriction each division inherits. In the middle right, list the five Pythagorean conjugate pairs and what each product collapses to. In the bottom left, draw the quadrant sign chart for all six functions, and mark clearly which two functions have a different sign pattern from cosine and sine. In the bottom right, verify completely that the tangent times the cosecant equals the secant. Finally, circle the two functions on your page that are never undefined, and write one sentence saying why they are the exception.
The circled pair should be cosine and sine, and the reason is that they are coordinates rather than quotients. Every other function on the page is a fraction built from them, and a fraction can have a zero denominator.
Recap
Five things, and the second one is the habit that prevents the most common error in this section.
| If the question says | Your first move is |
|---|---|
| Evaluate this function at this angle | Find the cosine and sine, then check the denominator |
| Solve an equation in secant or cosecant | Reciprocate to get a cosine or sine equation |
| Solve an equation in tangent or cotangent | Reference angle from the size, opposite quadrants from the sign |
| Simplify this expression | Look for a Pythagorean identity in an alternate form |
| Verify this identity | Start on the more complicated side; convert to cos and sin if stuck |
The next lesson takes these six functions off the unit circle, exactly as Lesson 10.2c did for cosine and sine, and lands them in a right triangle of any size. That is where the angle of elevation problems live, and where the whole apparatus finally starts measuring real distances.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 744-752 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Trigonometry — $55/session, free consultation.