Precalculus · G.SRT.C.6

Secant, cosecant and cotangent ratios

Sine, cosine and tangent each divide one side of a right triangle by another. What happens when the same two sides are divided the other way up, and why would anyone want those numbers?

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Why this exists

A right triangle has three sides, and two sides can be divided in two orders. Sine, cosine and tangent use three of the six possible quotients. The other three are their reciprocals, and they get names too: secant, cosecant and cotangent.

They carry no new information, since each is 1 divided by a ratio already known. They exist because the side being found is often on top. A crane operator who knows the horizontal reach of the boom and its angle wants the boom’s length, which is the hypotenuse over the reach. That quotient is the secant of the angle, and with a name it can be computed and written in one step instead of as a fraction with a cosine in its denominator. The same three ratios turn up again in the identities of trigonometry and, later, in calculus, so they are worth knowing by name.

Definition

Take an acute angle A in a right triangle and name the sides from A, as in opposite, adjacent and hypotenuse. The three reciprocal trigonometric ratios of A are

\begin{aligned} \sec A &= \frac{\text{hypotenuse}}{\text{adjacent}} \\ \csc A &= \frac{\text{hypotenuse}}{\text{opposite}} \\ \cot A &= \frac{\text{adjacent}}{\text{opposite}} \end{aligned}

read “the secant of A”, “the cosecant of A” and “the cotangent of A”. In the usual lettering of Figure 1, with the right angle at C,

\sec A = \frac{c}{b} \qquad \csc A = \frac{c}{a} \qquad \cot A = \frac{b}{a}

Figure 1From \angle A, a is opposite, b is adjacent and c is the hypotenuse. Each new ratio is a sine, cosine or tangent turned upside down.

Each is a ratio already known, turned upside down:

\begin{gathered} \sec A = \frac{1}{\cos A} \qquad \csc A = \frac{1}{\sin A} \\ \cot A = \frac{1}{\tan A} \end{gathered}

The pairing crosses the names: the secant goes with the cosine and the cosecant with the sine. Each pair has exactly one “co” in it.

Intuition

Hypotenuse on top

Sine and cosine put the hypotenuse on the bottom, so they say how much of the boom becomes height or reach. Secant and cosecant put it on top, so they say how much boom is needed: for each foot of reach, the boom must be \sec A feet long; for each foot of height, \csc A feet long. The cotangent is run over rise, the reach gained for each foot of height.

Why “secant”

Scale the triangle so the adjacent side is 1, as in Figure 2: put A at the center of a circle of radius 1, with the adjacent side AD a radius, and call the angle at A by the name \theta. The vertical line at D touches the circle, and “tangent” is Latin for touching: the piece DE of that line has length \displaystyle \frac{\text{opposite}}{\text{adjacent}} = \tan\theta. The hypotenuse AE runs out through the circle and cuts it, and “secant” is Latin for cutting: its length is \displaystyle \frac{\text{hypotenuse}}{\text{adjacent}} = \sec\theta. This is the same word as a secant line of a circle.

Figure 2With the adjacent side equal to 1, the touching line gives \tan\theta and the cutting line gives \sec\theta.

As \theta opens toward 90^\circ, the point E climbs the touching line without limit, so \sec\theta grows without limit. As \theta closes toward 0^\circ, E comes down to D and \sec\theta comes down toward 1.

Results

Reciprocals

For every acute angle A,

\begin{gathered} \sin A \csc A = 1 \qquad \cos A \sec A = 1 \\ \tan A \cot A = 1 \end{gathered}

and the cotangent is a quotient of the sine and cosine:

\cot A = \frac{\cos A}{\sin A}

Complementary angles

If A and B are the two acute angles of a right triangle, so that A + B = 90^\circ, then

\sec A = \csc B \qquad \cot A = \tan B

Two Pythagorean forms

For every acute angle A,

\begin{gathered} 1 + \tan^2 A = \sec^2 A \\ 1 + \cot^2 A = \csc^2 A \end{gathered}

Sizes

For every acute angle A, \sec A > 1 and \csc A > 1, while \cot A can be any positive number. As A grows, \sec A grows, and \csc A and \cot A shrink.

Why it is true

Reciprocals. \displaystyle \frac{c}{b} is the reciprocal of \displaystyle \frac{b}{c}, and likewise for the other two pairs, so each product is 1. The cotangent is \displaystyle \frac{b}{a} = \frac{b/c}{a/c} = \frac{\cos A}{\sin A}, dividing top and bottom by c, which is not 0.

Complementary angles. Seen from B, the side b is opposite and a is adjacent. So \displaystyle \csc B = \frac{c}{b} = \sec A and \displaystyle \tan B = \frac{b}{a} = \cot A. This is where the “co” comes from: the cosecant of an angle is the secant of its complement, and the cotangent is the tangent of its complement, just as the cosine is the sine of the complement.

Pythagorean forms. Start from a^2 + b^2 = c^2, the Pythagorean theorem. Dividing every term by b^2, and then instead by a^2, gives

\begin{gathered} \left(\frac{a}{b}\right)^2 + 1 = \left(\frac{c}{b}\right)^2 \\ 1 + \left(\frac{b}{a}\right)^2 = \left(\frac{c}{a}\right)^2 \end{gathered}

which are \tan^2 A + 1 = \sec^2 A and 1 + \cot^2 A = \csc^2 A.

Sizes. The sine and cosine of an acute angle lie strictly between 0 and 1, and 1 divided by such a number is greater than 1. The tangent can be any positive number, so its reciprocal can too. Taking reciprocals of positive numbers reverses their order, and as A grows the cosine shrinks while the sine and tangent grow; so the secant grows and the cosecant and cotangent shrink.

Worked examples

Example 1

Six ratios off the crane

In Figure 3 a crane’s boom is 65 feet long from the pivot P to the tip T. The tip is 33 feet above the pivot and 56 feet out from it, and the boom makes angle \theta with the horizontal. Find all six ratios of \theta, then the secant, cosecant and cotangent of the angle at T.

Figure 3The boom, its height and its reach make a right triangle. For Example 1.

Solution

From \theta at P, the hypotenuse is the boom, 65; the opposite side is the height, 33; the adjacent side is the reach, 56. So

\begin{aligned} \sin\theta &= \tfrac{33}{65} & \cos\theta &= \tfrac{56}{65} & \tan\theta &= \tfrac{33}{56} \\ \csc\theta &= \tfrac{65}{33} & \sec\theta &= \tfrac{65}{56} & \cot\theta &= \tfrac{56}{33} \end{aligned}

The bottom row is the top row turned over, in the crossed pairs: \sec with \cos, \csc with \sin. At T the opposite side is the reach, 56, and the adjacent side is the height, 33:

\sec T = \tfrac{65}{33} \quad \csc T = \tfrac{65}{56} \quad \cot T = \tfrac{33}{56}

so \sec T = \csc\theta and \cot T = \tan\theta, as the complementary-angle result says.

Check. 33^2 + 56^2 = 1089 + 3136 = 4225 = 65^2. Also

\begin{aligned} 1 + \tan^2\theta &= \frac{3136 + 1089}{3136} \\ &= \frac{4225}{3136} = \sec^2\theta \end{aligned}

Example 2

A leg to find first

In Figure 4, \triangle GHJ has its right angle at H, hypotenuse GJ = 11 and HJ = 5. Find \csc G, \sec G and \cot G.

Figure 4One leg is missing. For Example 2.

Solution

From G, HJ = 5 is opposite and GH, the adjacent side, is missing. By the Pythagorean theorem, \displaystyle GH = \sqrt{11^2 - 5^2} = \sqrt{96} = 4\sqrt{6}, the positive root because GH is a length. Then

\begin{aligned} \csc G &= \frac{11}{5} \\ \sec G &= \frac{11}{4\sqrt{6}} = \frac{11\sqrt{6}}{24} \\ \cot G &= \frac{4\sqrt{6}}{5} \end{aligned}

The secant was rationalized by multiplying top and bottom by \displaystyle \sqrt{6}.

Check. \sec G \approx 1.12 and \csc G = 2.2 are both greater than 1, as they must be.

Example 3

From the secant to the rest

An acute angle \theta has \displaystyle \sec\theta = \frac{41}{9}. Find its other five ratios.

Solution

The secant is hypotenuse over adjacent, so take a triangle with adjacent side 9 and hypotenuse 41; any similar one gives the same ratios. The opposite side is \displaystyle \sqrt{41^2 - 9^2} = \sqrt{1600} = 40. So

\begin{aligned} \cos\theta &= \tfrac{9}{41} & \sin\theta &= \tfrac{40}{41} & \tan\theta &= \tfrac{40}{9} \\ & & \csc\theta &= \tfrac{41}{40} & \cot\theta &= \tfrac{9}{40} \end{aligned}

Check. With the first Pythagorean form,

\begin{aligned} 1 + \left(\frac{40}{9}\right)^2 &= \frac{81 + 1600}{81} \\ &= \frac{1681}{81} = \left(\frac{41}{9}\right)^2 \end{aligned}

Example 4

The boom length from a calculator

A crane’s boom makes a 62^\circ angle with the horizontal and reaches 84 feet out from the pivot. Find \sec 62^\circ, \csc 62^\circ and \cot 62^\circ to four decimal places, and then the length of the boom.

Solution

Most calculators have no secant key, so use the reciprocals, in degree mode:

\begin{aligned} \sec 62^\circ &= \frac{1}{\cos 62^\circ} \approx 2.1301 \\ \csc 62^\circ &= \frac{1}{\sin 62^\circ} \approx 1.1326 \\ \cot 62^\circ &= \frac{1}{\tan 62^\circ} \approx 0.5317 \end{aligned}

The boom is the hypotenuse and the reach is adjacent to the 62^\circ angle, so \displaystyle \frac{\text{boom}}{84} = \sec 62^\circ and

\text{boom} = 84 \sec 62^\circ \approx 178.9 \text{ feet}

The unknown is on top of the secant, so one multiplication finishes the job.

Example 5

Exact values at the special angles

Find \sec, \csc and \cot of 30^\circ, 45^\circ and 60^\circ exactly.

Solution

Flip the values from the special right triangles, rationalizing where a root lands on the bottom:

angle	\sec	\csc	\cot
30^\circ	\displaystyle \frac{2\sqrt{3}}{3}	2	\displaystyle \sqrt{3}
45^\circ	\displaystyle \sqrt{2}	\displaystyle \sqrt{2}	1
60^\circ	2	\displaystyle \frac{2\sqrt{3}}{3}	\displaystyle \frac{\sqrt{3}}{3}

For example \displaystyle \sec 30^\circ = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}. The 30^\circ and 60^\circ rows trade their secant and cosecant, and the 45^\circ row matches itself, because 45^\circ is its own complement.

Example 6

Proving an identity

Show that \csc\theta - \sin\theta = \cos\theta \cot\theta for every acute angle \theta.

Solution

Write the left side in sines and cosines and combine over one denominator:

\begin{aligned} \csc\theta - \sin\theta &= \frac{1}{\sin\theta} - \sin\theta \\ &= \frac{1 - \sin^2\theta}{\sin\theta} = \frac{\cos^2\theta}{\sin\theta} \\ &= \cos\theta \cdot \frac{\cos\theta}{\sin\theta} \\ &= \cos\theta \cot\theta \end{aligned}

The second line uses \sin^2\theta + \cos^2\theta = 1, and the last uses \displaystyle \cot\theta = \frac{\cos\theta}{\sin\theta}.

Check. With the crane’s \theta from Example 1:

\begin{aligned} \frac{65}{33} - \frac{33}{65} &= \frac{4225 - 1089}{2145} = \frac{3136}{2145} \\ \frac{56}{65} \cdot \frac{56}{33} &= \frac{3136}{2145} \end{aligned}

Example 7

An equation from complements

Find x if \sec(2x + 10)^\circ = \csc(3x - 5)^\circ and both angles are acute.

Solution

The cosecant of an angle is the secant of its complement, so

\begin{aligned} \csc(3x - 5)^\circ &= \sec\bigl(90 - (3x - 5)\bigr)^\circ \\ &= \sec(95 - 3x)^\circ \end{aligned}

The secant grows as an acute angle grows, so two different acute angles never share a secant, and

\begin{aligned} 2x + 10 &= 95 - 3x \\ 5x &= 85 \\ x &= 17 \end{aligned}

Check. The angles are 2(17) + 10 = 44 and 3(17) - 5 = 46 degrees. Both are acute and they add to 90^\circ.

Boundaries

Acute angles only, for now. These definitions need a right triangle, so they give values only for 0^\circ < A < 90^\circ, where no side is 0 and every quotient exists. On the unit circle the ratios extend to all angles, and there a zero coordinate leaves some of them undefined, such as \sec 90^\circ.

Reciprocal is not inverse. \sec A is \displaystyle \frac{1}{\cos A}, a number. The calculator key \cos^{-1} does something else: it takes a ratio back to an angle, as in the inverse ratios. Only (\cos A)^{-1}, with the parentheses, means \sec A.

Degree mode. As with sine and cosine, the angles here are in degrees, and a calculator in radian mode reads \cos 62 as a different angle.

A ratio needs its angle. In one triangle \displaystyle \frac{c}{b} is both \sec A and \csc B.

Where this goes next

The reciprocal and quotient relations here, with \displaystyle \tan A = \frac{\sin A}{\cos A} and \displaystyle \cot A = \frac{\cos A}{\sin A}, are collected as the reciprocal and quotient identities. The complementary pairs grow into the cofunction identities, and 1 + \tan^2 A = \sec^2 A joins \sin^2 A + \cos^2 A = 1 among the Pythagorean identities. In problems, a secant or cosecant puts an unknown hypotenuse on top, which is often the quickest way to solve a right triangle. On the unit circle all six ratios are defined for every angle where their denominators are not 0, which is the subject of secant, cosecant and cotangent on the unit circle.

Exercises

Give exact answers unless a decimal is asked for. Every angle named \theta, A or B is acute.

Reading the ratios off a triangle

Triangle DEF has a right angle at E, with DE = 20, EF = 99 and DF = 101.

Find \sec D, \csc D and \cot D.
Find \sec F, \csc F and \cot F.
A right triangle has hypotenuse 10 and one leg 3, and A is the angle opposite that leg. Find \csc A, \sec A and \cot A.

One ratio given

\displaystyle \sec\theta = \frac{37}{12}. Find the other five ratios of \theta.
\displaystyle \cot\theta = \frac{2}{3}. Find \csc\theta and \sec\theta.
\csc A = 3. Find \sec A and \cot A.
\displaystyle \cos\theta = \frac{2}{5}. Find \sec\theta, \csc\theta and \cot\theta.

Calculators and cranes

Find \sec 25^\circ, \csc 25^\circ and \cot 25^\circ to four decimal places.
Bauyrzhan’s crane holds its boom at 58^\circ above horizontal with the tip 90 feet above the pivot. Use a cosecant to find the boom’s length to the nearest tenth of a foot.
A boom at 40^\circ has its tip 75 feet above the pivot. Use a cotangent to find the horizontal reach to the nearest tenth of a foot.
Mpho’s crane has a 150-foot boom with a horizontal reach of 90 feet. Find \sec\theta, \csc\theta and \cot\theta for the boom’s angle \theta, and the height of the tip above the pivot.

Exact values and identities

Find \sec^2 45^\circ + \cot^2 30^\circ exactly.
Find \csc 30^\circ \cdot \sec 60^\circ - \cot 45^\circ exactly.
\displaystyle \tan\theta = \frac{1}{2}. Use 1 + \tan^2\theta = \sec^2\theta to find \sec\theta.
Show that \sec\theta \cot\theta = \csc\theta.
Show that \tan\theta + \cot\theta = \sec\theta \csc\theta.

Complementary angles

\sec 17^\circ \approx 1.0457. Without a calculator, give \csc 73^\circ.
Find x if \cot(x + 12)^\circ = \tan(2x + 3)^\circ and both angles are acute.

Reasoning

Explain why \sec A > 1 and \csc A > 1 for every acute angle A, but \cot A can be as small as 0.001.
Prove 1 + \cot^2 A = \csc^2 A from \sin^2 A + \cos^2 A = 1.
Konstancja raises a crane’s boom with the reach held fixed. Explain what happens to the boom length the crane needs, using \sec\theta.
Explain why the secant of a boom’s angle does not depend on how long the boom is.

Answers

\displaystyle \sec D = \frac{101}{20}, \displaystyle \csc D = \frac{101}{99}, \displaystyle \cot D = \frac{20}{99}. From D, the adjacent side is DE = 20 and the opposite side is EF = 99.
\displaystyle \sec F = \frac{101}{99}, \displaystyle \csc F = \frac{101}{20}, \displaystyle \cot F = \frac{99}{20}. From F the two legs trade roles.
\displaystyle \csc A = \frac{10}{3}, \displaystyle \sec A = \frac{10\sqrt{91}}{91}, \displaystyle \cot A = \frac{\sqrt{91}}{3}. The other leg is \displaystyle \sqrt{100 - 9} = \sqrt{91}.
\displaystyle \cos\theta = \frac{12}{37}, \displaystyle \sin\theta = \frac{35}{37}, \displaystyle \tan\theta = \frac{35}{12}, \displaystyle \csc\theta = \frac{37}{35}, \displaystyle \cot\theta = \frac{12}{35}. The opposite side is \displaystyle \sqrt{1369 - 144} = \sqrt{1225} = 35.
\displaystyle \csc\theta = \frac{\sqrt{13}}{3}, \displaystyle \sec\theta = \frac{\sqrt{13}}{2}. Take adjacent 2 and opposite 3; the hypotenuse is \displaystyle \sqrt{4 + 9} = \sqrt{13}.
\displaystyle \sec A = \frac{3\sqrt{2}}{4}, \displaystyle \cot A = 2\sqrt{2}. Take opposite 1 and hypotenuse 3; the adjacent side is \displaystyle \sqrt{9 - 1} = 2\sqrt{2}, so \displaystyle \sec A = \frac{3}{2\sqrt{2}}.
\displaystyle \sec\theta = \frac{5}{2}, \displaystyle \csc\theta = \frac{5\sqrt{21}}{21}, \displaystyle \cot\theta = \frac{2\sqrt{21}}{21}. Take adjacent 2 and hypotenuse 5; the opposite side is \displaystyle \sqrt{21}, so \displaystyle \csc\theta = \frac{5}{\sqrt{21}} and \displaystyle \cot\theta = \frac{2}{\sqrt{21}}.
\sec 25^\circ \approx 1.1034, \csc 25^\circ \approx 2.3662, \cot 25^\circ \approx 2.1445, as \displaystyle \frac{1}{\cos 25^\circ}, \displaystyle \frac{1}{\sin 25^\circ} and \displaystyle \frac{1}{\tan 25^\circ}.
About 106.1 feet. The height is opposite the angle, so \displaystyle \frac{\text{boom}}{90} = \csc 58^\circ and the boom is 90 \csc 58^\circ \approx 90(1.1792).
About 89.4 feet. The cotangent is reach over height, so the reach is 75 \cot 40^\circ \approx 75(1.1918).
\displaystyle \sec\theta = \frac{5}{3}, \displaystyle \csc\theta = \frac{5}{4}, \displaystyle \cot\theta = \frac{3}{4}; the tip is 120 feet up. The height is \displaystyle \sqrt{150^2 - 90^2} = \sqrt{14{,}400} = 120, and \displaystyle \sec\theta = \frac{150}{90}, \displaystyle \csc\theta = \frac{150}{120}, \displaystyle \cot\theta = \frac{90}{120}.
5. \displaystyle \sec^2 45^\circ = (\sqrt{2})^2 = 2 and \displaystyle \cot^2 30^\circ = (\sqrt{3})^2 = 3.
3. It is 2 \cdot 2 - 1.
\displaystyle \sec\theta = \frac{\sqrt{5}}{2}. \displaystyle \sec^2\theta = 1 + \frac{1}{4} = \frac{5}{4}, and the secant of an acute angle is positive.
Write both in sines and cosines and cancel \cos\theta, which is not 0: \displaystyle \sec\theta \cot\theta = \frac{1}{\cos\theta} \cdot \frac{\cos\theta}{\sin\theta}, which is \displaystyle \frac{1}{\sin\theta} = \csc\theta.
Over a common denominator, \displaystyle \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} is \displaystyle \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta}, and the top is 1, leaving \displaystyle \frac{1}{\sin\theta\cos\theta} = \sec\theta \csc\theta.
\csc 73^\circ \approx 1.0457. Since 17 + 73 = 90, \csc 73^\circ = \sec 17^\circ.
x = 25; the angles are 37^\circ and 53^\circ. The tangent is the cotangent of the complement, so x + 12 = 90 - (2x + 3), giving 3x = 75.
The hypotenuse is longer than either leg, so hypotenuse over a leg is greater than 1. The cotangent is one leg over the other, and legs 1 and 1000 give \cot A = 0.001 at the angle opposite the long leg.
Divide \sin^2 A + \cos^2 A = 1 by \sin^2 A, which is not 0: \displaystyle 1 + \frac{\cos^2 A}{\sin^2 A} = \frac{1}{\sin^2 A}, which is 1 + \cot^2 A = \csc^2 A.
The boom length is the reach times \sec\theta. As \theta grows, \cos\theta shrinks, so \sec\theta grows, and without limit near 90^\circ: a steeper boom with the same reach must be longer, and no boom can reach a nonzero distance standing straight up.
Any two booms at the same angle, with their heights and reaches, form right triangles with a common acute angle, which are similar by AA similarity. Their sides are in proportion, so boom over reach is the same number for both.