Precalculus

Reciprocal and quotient identities

Secant, cosecant, tangent and cotangent are all built from the two coordinates that sine and cosine already name. Can every one of them be written in sine and cosine alone, and when is the rewritten form allowed to replace the original?

You need

Secant, cosecant and cotangent on the unit circlePrecalculus What an identity isPrecalculus

Unlocks

The Pythagorean identitiesPrecalculus

Where you are 13 in Trigonometric identities

Trigonometry7

Trigonometric identities13

Cofunction identities What an identity is

Reciprocal and quotient identities

The Pythagorean identities Cosine of a sum or difference ▾ Show all 13 (8 more)

Why this exists

The lesson on the six functions on the unit circle defined secant, cosecant, tangent and cotangent as fractions built from the terminal point (x, y), and sine and cosine as the point’s coordinates. Every one of the six is therefore made from the same two numbers. Saying so as equations between the functions themselves, such as \displaystyle \tan t = \frac{\sin t}{\cos t}, turns a definition into a tool: any expression in the six functions can be rewritten in sine and cosine, and two expressions that look different can be compared on equal terms.

A tool used to replace one expression by another has to come with the set where the replacement is honest. That is what makes these statements identities rather than loose abbreviations: each holds on its common domain, and the common domain is part of the statement.

Definition

The reciprocal identities say that three of the functions are one over the other three. The quotient identities say that tangent and cotangent are quotients of sine and cosine. For every angle t at which both sides are defined,

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

\tan t = \frac{\sin t}{\cos t} \qquad \cot t = \frac{\cos t}{\sin t}

These five, with the Pythagorean identities that come next, are often called the fundamental identities. In right triangles the same reciprocal pairs appear as ratios of sides; the identities extend them to every angle at which the functions exist.

Intuition

Think of \sin t and \cos t as the two coordinates and the other four names as shorthand for fractions made from them. Rewriting in sine and cosine undoes the shorthand. Once it is undone, every denominator is in plain sight, so the places where an expression has no value can be read off directly: a factor \sec t hides a \cos t in a denominator, and a factor \csc t hides a \sin t.

That is also why simplifying can change the domain. When \displaystyle \frac{\sec x}{\csc x} is rewritten as \displaystyle \frac{\sin x}{\cos x}, the hidden \sin x in the cosecant’s denominator moves to the top and no longer forbids anything. The short form is defined at more inputs than the long one, and the two agree only where both exist.

Results

Reciprocal and quotient identities. With k standing for any integer, each identity holds exactly on the set shown, its common domain.

Identity	Holds for
\displaystyle \csc t = \frac{1}{\sin t}	t \neq k\pi
\displaystyle \sec t = \frac{1}{\cos t}	\displaystyle t \neq \frac{\pi}{2} + k\pi
\displaystyle \cot t = \frac{1}{\tan t}	\displaystyle t \neq \frac{k\pi}{2}
\displaystyle \tan t = \frac{\sin t}{\cos t}	\displaystyle t \neq \frac{\pi}{2} + k\pi
\displaystyle \cot t = \frac{\cos t}{\sin t}	t \neq k\pi

Turned around, and as products. On the same sets,

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

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

so \displaystyle \sin t = \frac{1}{\csc t} and \sin t \csc t = 1 hold for t \neq k\pi, the cosine forms for \displaystyle t \neq \frac{\pi}{2} + k\pi, and the tangent forms for \displaystyle t \neq \frac{k\pi}{2}.

Everything in sine and cosine. Any expression built from the six functions of t by adding, multiplying and dividing equals an expression in \sin t and \cos t alone, at every t where the original is defined.

Why it is true

Let t have terminal point (x, y) on the unit circle, so \sin t = y and \cos t = x, and \displaystyle \csc t = \frac{1}{y}, \displaystyle \sec t = \frac{1}{x}, \displaystyle \tan t = \frac{y}{x}, \displaystyle \cot t = \frac{x}{y}.

The cosecant, secant and both quotient identities are these definitions with y and x renamed, so each side exists under the same condition (y \neq 0 or x \neq 0) and they agree wherever they exist. The condition y \neq 0 means t \neq k\pi, and x \neq 0 means \displaystyle t \neq \frac{\pi}{2} + k\pi.

For \displaystyle \cot t = \frac{1}{\tan t}, the right side needs \tan t to exist (x \neq 0) and to be nonzero (y \neq 0), and then \displaystyle \frac{1}{y/x} = \frac{x}{y} = \cot t. Both coordinates nonzero means t is not a multiple of \displaystyle \frac{\pi}{2}. The turned-around forms are the same fractions inverted, and each product is a nonzero number times its own reciprocal.

The last result follows by replacing each of the four functions by its fraction in \sin t and \cos t: every value in the original was defined, so every replacement is allowed.

Worked examples

Example 1

Secant over cosecant

Write \displaystyle \frac{\sec x}{\csc x} as a single trigonometric function, and state the set on which the two are equal.

Solution

First find where the left side exists. The secant needs \cos x \neq 0, the cosecant needs \sin x \neq 0, and a cosecant is never 0, so dividing by it adds no further condition. The left side is defined for \displaystyle x \neq \frac{k\pi}{2}, k any integer.

On that set replace each function by its reciprocal form, and divide by multiplying by the reciprocal:

\begin{aligned} \frac{\sec x}{\csc x} &= \frac{1}{\cos x} \cdot \frac{\sin x}{1} \\ &= \frac{\sin x}{\cos x} = \tan x \end{aligned}

The right side, \tan x, is defined on the larger set \displaystyle x \neq \frac{\pi}{2} + k\pi, so the common domain is the left side’s. The identity \displaystyle \frac{\sec x}{\csc x} = \tan x holds for \displaystyle x \neq \frac{k\pi}{2}. At each multiple of \pi the tangent is 0 while the left side has no value, so the graph of the left side is the tangent graph with holes at the points (k\pi, 0).

Figure 1The graph of \displaystyle y = \frac{\sec x}{\csc x} is the tangent graph with a hole at every multiple of \pi, where \tan x = 0 but the cosecant has no value.

Example 2

Cotangent times secant

Show that \cot x \sec x = \csc x is an identity, and find the holes in the graph of y = \cot x \sec x for -2\pi \le x \le 2\pi.

Solution

The cotangent needs \sin x \neq 0 and the secant needs \cos x \neq 0, so the left side is defined for \displaystyle x \neq \frac{k\pi}{2}. On that set,

\begin{aligned} \cot x \sec x &= \frac{\cos x}{\sin x} \cdot \frac{1}{\cos x} \\ &= \frac{1}{\sin x} = \csc x \end{aligned}

where the cosines cancel because \cos x \neq 0. The cosecant is defined on x \neq k\pi, a larger set, so the identity holds for \displaystyle x \neq \frac{k\pi}{2}.

The holes are where the cosecant exists but the left side does not: \displaystyle x = \frac{\pi}{2} + k\pi, where \displaystyle \csc x = \frac{1}{\sin x} = \pm 1. In the window they are \displaystyle \left(-\frac{3\pi}{2}, 1\right), \displaystyle \left(-\frac{\pi}{2}, -1\right), \displaystyle \left(\frac{\pi}{2}, 1\right) and \displaystyle \left(\frac{3\pi}{2}, -1\right), the turning points of the cosecant’s branches.

Figure 2The graph of y = \cot x \sec x is the cosecant graph with its turning points removed, where \cos x = 0 and the secant has no value.

Example 3

Values a calculator has no key for

Kerttu’s calculator has keys for \sin, \cos and \tan only. Find \sec 4, \csc 4 and \cot 4, with 4 in radians, to four decimal places.

Solution

The reciprocal identities turn each value into one the calculator has. Since \displaystyle \pi < 4 < \frac{3\pi}{2}, the angle is in Quadrant III, where no coordinate is 0, so all three exist:

\begin{aligned} \sec 4 &= \frac{1}{\cos 4} \approx \frac{1}{-0.653644} \approx -1.5299 \\ \csc 4 &= \frac{1}{\sin 4} \approx \frac{1}{-0.756802} \approx -1.3213 \\ \cot 4 &= \frac{1}{\tan 4} \approx \frac{1}{1.157821} \approx 0.8637 \end{aligned}

The key marked \cos^{-1} does something else: it gives the angle whose cosine is a given number, and \cos^{-1} 4 does not exist, since no cosine is 4. Each reciprocal keeps the sign of its partner, as it must: in Quadrant III sine and cosine are negative and tangent is positive.

Example 4

Rewrite first, then evaluate

Find the exact value of \displaystyle \csc\frac{5\pi}{3}\tan\frac{5\pi}{3}.

Solution

The angle \displaystyle \frac{5\pi}{3} is not a multiple of \displaystyle \frac{\pi}{2}, so both functions are defined and the identities may be used there. As in Example 1,

\begin{aligned} \csc t \tan t &= \frac{1}{\sin t} \cdot \frac{\sin t}{\cos t} \\ &= \frac{1}{\cos t} = \sec t \end{aligned}

so only one value is needed. The terminal point of \displaystyle \frac{5\pi}{3} is \displaystyle \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right), so \displaystyle \cos\frac{5\pi}{3} = \frac{1}{2} and the product is 2.

Check. Directly, \displaystyle \csc\frac{5\pi}{3} = -\frac{2}{\sqrt{3}} and \displaystyle \tan\frac{5\pi}{3} = -\sqrt{3}, and \displaystyle \left(-\frac{2}{\sqrt{3}}\right)\left(-\sqrt{3}\right) = 2.

Example 5

Sine and cosine from tangent and secant

\tan\theta = 3 and \displaystyle \sec\theta = -\sqrt{10}. Find \sin\theta, \cos\theta, \cot\theta and \csc\theta.

Solution

Cosine is the reciprocal of secant: \displaystyle \cos\theta = -\frac{1}{\sqrt{10}} = -\frac{\sqrt{10}}{10}. The quotient identity \displaystyle \tan\theta = \frac{\sin\theta}{\cos\theta}, multiplied through by the nonzero \cos\theta, gives \sin\theta = \tan\theta\cos\theta:

\sin\theta = 3 \cdot \left(-\frac{\sqrt{10}}{10}\right) = -\frac{3\sqrt{10}}{10}

The other two are reciprocals: \displaystyle \cot\theta = \frac{1}{3} and \displaystyle \csc\theta = -\frac{10}{3\sqrt{10}} = -\frac{\sqrt{10}}{3}. Both sine and cosine are negative, so \theta is in Quadrant III.

Check. The point (\cos\theta, \sin\theta) must be on the unit circle: \displaystyle \frac{10}{100} + \frac{90}{100} = 1.

Example 6

A quotient of sums

Ilmari claims that \displaystyle \frac{1 + \tan x}{1 + \cot x} = \tan x. Find the common domain and decide whether he is right.

Solution

The left side needs \cos x \neq 0 for the tangent, \sin x \neq 0 for the cotangent, and a nonzero denominator, \cot x \neq -1. The cotangent is -1 at \displaystyle x = \frac{3\pi}{4} + k\pi. The right side adds no new condition, so the common domain is \displaystyle x \neq \frac{k\pi}{2} together with \displaystyle x \neq \frac{3\pi}{4} + k\pi.

On that set write everything in sine and cosine and put each sum over one denominator:

\begin{aligned} \frac{1 + \tan x}{1 + \cot x} &= \frac{(\cos x + \sin x)/\cos x}{(\sin x + \cos x)/\sin x} \\ &= \frac{\cos x + \sin x}{\cos x} \cdot \frac{\sin x}{\sin x + \cos x} \\ &= \frac{\sin x}{\cos x} = \tan x \end{aligned}

The factor \sin x + \cos x may be cancelled because it is not 0: it equals \sin x\,(1 + \cot x), a product of two nonzero numbers. So Ilmari is right on the common domain. The graph of the left side is the tangent graph with holes at (k\pi, 0) and at \displaystyle \left(\frac{3\pi}{4} + k\pi, -1\right); at \displaystyle \frac{3\pi}{4} the left side is the form \displaystyle \frac{0}{0}.

Example 7

An equation with a domain to respect

Solve \tan x = 2\sin x for 0 \le x < 2\pi.

Solution

Any solution must have \cos x \neq 0, or the tangent would not exist. With that condition, \displaystyle \tan x = \frac{\sin x}{\cos x}, and multiplying both sides by the nonzero \cos x gives an equation with the same solutions:

\begin{aligned} \sin x &= 2\sin x\cos x \\ \sin x\,(1 - 2\cos x) &= 0 \end{aligned}

A product is 0 when a factor is. Either \sin x = 0, giving x = 0 or x = \pi, or \displaystyle \cos x = \frac{1}{2}, giving \displaystyle x = \frac{\pi}{3} or \displaystyle x = \frac{5\pi}{3}. None of the four has \cos x = 0, so all four stand:

x = 0, \quad \frac{\pi}{3}, \quad \pi, \quad \frac{5\pi}{3}

Dividing both sides by \sin x instead of factoring would have lost x = 0 and x = \pi.

Boundaries

The turned-around and product forms have smaller domains. \displaystyle \cot t = \frac{\cos t}{\sin t} holds wherever the cotangent exists, but \displaystyle \cot t = \frac{1}{\tan t} fails at \displaystyle t = \frac{\pi}{2} + k\pi, where \cot t = 0 and the tangent does not exist. Likewise \displaystyle \sin t = \frac{1}{\csc t} says nothing at t = k\pi, where \sin t = 0 is a perfectly good value.

Rewriting can widen the domain. A simplified expression may exist at inputs the original excluded, as \tan x does at x = 0 in Example 1. When an identity is used to solve an equation, an input excluded at the start stays excluded, even if the rewritten equation accepts it.

The exponent -1 means two things. \displaystyle (\cos x)^{-1} = \frac{1}{\cos x} = \sec x, but \cos^{-1} x is the inverse cosine, an angle. They differ: \displaystyle \cos^{-1}\frac{1}{2} = \frac{\pi}{3}, while \displaystyle \left(\cos\frac{1}{2}\right)^{-1} \approx 1.1395.

Where this goes next

With every function written in sine and cosine, the Pythagorean identities relate \sin t and \cos t to each other, which is what Example 5 used as a check; together they let one value and a sign give all six. The Cofunction identities pair each function with its partner of the complementary angle, which is where the “co” in cosine, cotangent and cosecant comes from. The habit of Examples 1, 2 and 6, rewrite in sine and cosine, then combine, is the main move in simplifying expressions and in verifying identities, and Example 7 is the first of many trigonometric equations. The holes and asymptotes in Figures 1 and 2 come from the gaps described in Graphing tangent and the reciprocal functions.

Exercises

Rewrite in sine and cosine

Write each expression as a single trigonometric function or a number, and state the set on which the two are equal.

\tan x \csc x
\displaystyle \frac{\cot x}{\csc x}
\sin x \sec x
\displaystyle \frac{\csc x}{\sec x}
\cos x \tan x \csc x
\displaystyle \frac{\csc x}{\cot x}

Values

Find the exact value of \displaystyle \sec\frac{\pi}{3}\cot\frac{\pi}{3} by rewriting it first.
Find the exact value of \displaystyle \frac{\tan(7\pi/6)}{\sec(7\pi/6)} by rewriting it first.
\sin\theta = 0.28 and \cos\theta = -0.96. Find the other four values as fractions.
\displaystyle \cot\theta = -\frac{2}{3} and \displaystyle \csc\theta = -\frac{\sqrt{13}}{3}. Find \sin\theta, \cos\theta, \tan\theta and \sec\theta exactly.
Using only the \sin, \cos and \tan keys, find \sec 1.2, \csc 1.2 and \cot 1.2, with 1.2 in radians, to four decimal places.

Identities and their holes

\tan x \cot x = 1 is an identity. State its common domain and list the holes in the graph of y = \tan x \cot x for 0 \le x \le 2\pi.
Show that \displaystyle \frac{1}{\sec x \csc x} = \sin x \cos x on its common domain, and say where the graph of the left side has holes.
Show that \displaystyle \frac{\tan x + \sin x}{\sec x + 1} = \sin x on its common domain, and state that domain.

Equations

Solve each equation for 0 \le x < 2\pi.

\sec x = \csc x
\cot x = \csc x
\displaystyle \sin x \sec x = -\sqrt{3}

Reasoning

Explain why \sin t \csc t = 1 is an identity even though it fails at t = \pi.
Saini evaluates \sin\pi as \displaystyle \frac{1}{\csc\pi}. What goes wrong, and what is \sin\pi?
Kerttu says \displaystyle \cot x = \frac{1}{\tan x} and \displaystyle \cot x = \frac{\cos x}{\sin x} are the same identity. In what sense is she right, and in what sense not?
Explain why \tan t and \cot t have the same sign at every angle where both are defined.
Show that wherever \tan\theta and \sec\theta are both defined, \displaystyle \sin\theta = \frac{\tan\theta}{\sec\theta}.

Answers

\sec x, for \displaystyle x \neq \frac{k\pi}{2}. \displaystyle \frac{\sin x}{\cos x} \cdot \frac{1}{\sin x} = \frac{1}{\cos x}; the left side needs both \cos x \neq 0 and \sin x \neq 0.
\cos x, for x \neq k\pi. \displaystyle \frac{\cos x}{\sin x} \cdot \sin x = \cos x; both functions on the left need only \sin x \neq 0.
\tan x, for \displaystyle x \neq \frac{\pi}{2} + k\pi. \displaystyle \sin x \cdot \frac{1}{\cos x} = \frac{\sin x}{\cos x}, and both sides need exactly \cos x \neq 0, so the graphs match with no holes.
\cot x, for \displaystyle x \neq \frac{k\pi}{2}. \displaystyle \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x}; the secant on the left adds \cos x \neq 0.
1, for \displaystyle x \neq \frac{k\pi}{2}. \displaystyle \cos x \cdot \frac{\sin x}{\cos x} \cdot \frac{1}{\sin x} = 1.
\sec x, for \displaystyle x \neq \frac{k\pi}{2}. \displaystyle \frac{1}{\sin x} \cdot \frac{\sin x}{\cos x} = \frac{1}{\cos x}; dividing by \cot x needs \cot x \neq 0, that is \cos x \neq 0, on top of \sin x \neq 0.
\displaystyle \frac{2\sqrt{3}}{3}. \displaystyle \sec t \cot t = \frac{1}{\cos t} \cdot \frac{\cos t}{\sin t} = \csc t, and \displaystyle \csc\frac{\pi}{3} = \frac{2}{\sqrt{3}}.
\displaystyle -\frac{1}{2}. \displaystyle \frac{\tan t}{\sec t} = \frac{\sin t}{\cos t} \cdot \cos t = \sin t, and \displaystyle \sin\frac{7\pi}{6} = -\frac{1}{2}.
\displaystyle \tan\theta = -\frac{7}{24}, \displaystyle \cot\theta = -\frac{24}{7}, \displaystyle \sec\theta = -\frac{25}{24}, \displaystyle \csc\theta = \frac{25}{7}. Write \displaystyle 0.28 = \frac{7}{25} and \displaystyle -0.96 = -\frac{24}{25}.
\displaystyle \sin\theta = -\frac{3\sqrt{13}}{13}, \displaystyle \cos\theta = \frac{2\sqrt{13}}{13}, \displaystyle \tan\theta = -\frac{3}{2}, \displaystyle \sec\theta = \frac{\sqrt{13}}{2}. The sine is \displaystyle -\frac{3}{\sqrt{13}}, and \displaystyle \cos\theta = \cot\theta\sin\theta = \left(-\frac{2}{3}\right)\left(-\frac{3}{\sqrt{13}}\right) = \frac{2}{\sqrt{13}}.
\sec 1.2 \approx 2.7597, \csc 1.2 \approx 1.0729, \cot 1.2 \approx 0.3888, from \cos 1.2 \approx 0.362358, \sin 1.2 \approx 0.932039 and \tan 1.2 \approx 2.572152.
Common domain \displaystyle x \neq \frac{k\pi}{2}: the tangent needs \cos x \neq 0 and the cotangent \sin x \neq 0. The graph is the line y = 1 with holes at (0, 1), \displaystyle \left(\frac{\pi}{2}, 1\right), (\pi, 1), \displaystyle \left(\frac{3\pi}{2}, 1\right) and (2\pi, 1).
The left side needs \cos x \neq 0 and \sin x \neq 0, so its domain is \displaystyle x \neq \frac{k\pi}{2}. There \displaystyle \frac{1}{\sec x \csc x} = \frac{1}{\frac{1}{\cos x} \cdot \frac{1}{\sin x}} = \sin x \cos x. The right side is defined everywhere and is 0 at each multiple of \displaystyle \frac{\pi}{2}, so the holes are at \displaystyle \left(\frac{k\pi}{2}, 0\right).
Common domain \displaystyle x \neq \frac{\pi}{2} + k\pi and x \neq \pi + 2k\pi: the tangent and secant need \cos x \neq 0, and the denominator needs \sec x \neq -1, that is \cos x \neq -1. There the numerator is \displaystyle \frac{\sin x\,(1 + \cos x)}{\cos x} and the denominator is \displaystyle \frac{1 + \cos x}{\cos x}, so the quotient is \sin x, cancelling the nonzero \displaystyle \frac{1 + \cos x}{\cos x}.
\displaystyle x = \frac{\pi}{4} or \displaystyle x = \frac{5\pi}{4}. Both sides need \displaystyle x \neq \frac{k\pi}{2}; there \displaystyle \frac{1}{\cos x} = \frac{1}{\sin x} means \sin x = \cos x, so \tan x = 1.
No solution. Both sides need \sin x \neq 0; there \displaystyle \frac{\cos x}{\sin x} = \frac{1}{\sin x} means \cos x = 1, so x = 0, but \sin 0 = 0 excludes it.
\displaystyle x = \frac{2\pi}{3} or \displaystyle x = \frac{5\pi}{3}. The left side is \tan x wherever \cos x \neq 0, so \displaystyle \tan x = -\sqrt{3}; neither answer has \cos x = 0.
An identity has to hold only on its common domain. At t = \pi the cosecant has no value, so \pi is not tested; at every t \neq k\pi, \sin t is a nonzero number and \csc t is its reciprocal, so the product is 1.
\csc\pi does not exist, because \sin\pi = 0, so \displaystyle \frac{1}{\csc\pi} has no value. The identity \displaystyle \sin t = \frac{1}{\csc t} holds only for t \neq k\pi. The sine itself is fine: \sin\pi = 0, the y-coordinate of (-1, 0).
She is right that they agree wherever both right sides exist, at every \displaystyle x \neq \frac{k\pi}{2}. They are different statements because their domains differ: \displaystyle \frac{\cos x}{\sin x} also gives the right value 0 at \displaystyle x = \frac{\pi}{2} + k\pi, where \displaystyle \frac{1}{\tan x} has no value.
Where both are defined, \tan t \cot t = 1, which is positive. Two numbers with a positive product have the same sign.
Where both are defined, \cos\theta \neq 0 and \displaystyle \sec\theta = \frac{1}{\cos\theta} is never 0, so the quotient exists, and \displaystyle \frac{\tan\theta}{\sec\theta} = \frac{\sin\theta}{\cos\theta} \cdot \cos\theta = \sin\theta.