11.4 The Circular Functions: Tangent, Secant, Cosecant, and Cotangent

In section 11.2, we extended the notion of $\sin (θ)$ and $\cos (θ)$ from acute angles to any angles using the coordinate values of points on the Unit Circle. In total, there are six circular functions, as listed below.

Definition

The Circular Functions: Suppose an angle $θ$ is graphed in standard position.

Let $P (x, y)$ be the point of intersection of the terminal side of $P$ and the Unit Circle.

The sine of $θ$ , denoted $\sin (θ)$ , is defined by $\sin (θ) = y$ .
The cosine of $θ$ , denoted $\cos (θ)$ , is defined by $\cos (θ) = x$ .
The tangent of $θ$ , denoted $\tan (θ)$ , is defined by $\tan (θ) = \frac{y}{x}$ , provided $x not equal to 0$ .
The secant of $θ$ , denoted $\sec (θ)$ , is defined by $\sec (θ) = \frac{1}{x}$ , provided $x not equal to 0$ .
The cosecant of $θ$ , denoted $\csc (θ)$ , is defined by $\csc (θ) = \frac{1}{y}$ , provided $y not equal to 0$ .
The cotangent of $θ$ , denoted $\cot (θ)$ , is defined by $\cot (θ) = \frac{x}{y}$ , provided $y not equal to 0$ .

While we left the history of the name `sine' as an interesting research project in Section 11.2,we take a slight detour here to explain the origin of the names `tangent' and `secant.'

Consider the acute angle $θ$ in standard position sketched in the diagram below.

Coordinate-plane figure. — **Figure 11.134**

As usual, $P (x, y)$ denotes the point on the terminal side of $θ$ which lies on the Unit Circle, but we also consider the point $Q ⁢ open parenthesis 1, y prime close parenthesis$ , the point on the terminal side of $θ$ which lies on the vertical line $x equals 1$ .

The word `tangent' comes from the Latin meaning `to touch,' and for this reason, the line $x equals 1$ is called a tangent line to the Unit Circle since it intersects, or `touches', the circle at only one point, namely $open parenthesis 1, 0 close parenthesis$ .

Dropping perpendiculars from $P$ and $Q$ creates a pair of similar triangles $Δ O P A$ and $Δ O Q B$ . Hence the corresponding sides are proportional. We get $\frac{y^{'}}{y} = \frac{1}{x}$ which gives $y^{'} = \frac{y}{x} = \tan (θ)$ .

We have just shown that for acute angles $θ$ , $\tan (θ)$ is the $y$ -coordinate of the point on the terminal side of $θ$ which lies on the line $x equals 1$ which is tangent to the Unit Circle.

The word `secant' means `to cut', so a secant line is any line that `cuts through' a circle at two points.¹ The line containing the terminal side of $θ$ (not just the terminal side itself) is one such secant line since it intersects the Unit Circle in Quadrants I and III.

With the point $P$ lying on the Unit Circle, the length of the hypotenuse of $Δ O P A$ is $1$ . If we let $h$ denote the length of the hypotenuse of $Δ O Q B$ , we have from similar triangles that $\frac{h}{1} = \frac{1}{x}$ , or $h = \frac{1}{x} = \sec (θ)$ .

Hence for an acute angle $θ$ , $\sec (θ)$ is the length of the line segment which lies on the secant line determined by the terminal side of $θ$ and `cuts off' the tangent line $x equals 1$ .

As we mentioned in Definition 11.2, the `co' in `cosecant' and `cotangent' tie back to the concept of `co'mplementary angles and is explained in detail in Section 12.2.

Not only do these observations help explain the names of these functions, they serve as the basis for a fundamental inequality needed for Calculus which we'll explore in the Exercises.

Of the six circular functions, only sine and cosine are defined for all angles $θ$ . Since $x = \cos (θ)$ and $y = \sin (θ)$ in Definition 11.4, it is customary to rephrase the remaining four circular functions Definition 11.4 in terms of sine and cosine.

Theorem

Reciprocal and Quotient Identities:

$\sec (θ) = \frac{1}{\cos (θ)}$ , provided $cos ⁡ open parenthesis θ close parenthesis not equal to 0$ ; if $cos ⁡ open parenthesis θ close parenthesis equals 0$ , $\sec (θ)$ is undefined.
$\csc (θ) = \frac{1}{\sin (θ)}$ , provided $sin ⁡ open parenthesis θ close parenthesis not equal to 0$ ; if $sin ⁡ open parenthesis θ close parenthesis equals 0$ , $\csc (θ)$ is undefined.
$\tan (θ) = \frac{\sin (θ)}{\cos (θ)}$ , provided $cos ⁡ open parenthesis θ close parenthesis not equal to 0$ ; if $cos ⁡ open parenthesis θ close parenthesis equals 0$ , $\tan (θ)$ is undefined.
$\cot (θ) = \frac{\cos (θ)}{\sin (θ)}$ , provided $sin ⁡ open parenthesis θ close parenthesis not equal to 0$ ; if $sin ⁡ open parenthesis θ close parenthesis equals 0$ , $\cot (θ)$ is undefined.

We call the equations listed in Theorem 11.7 identities since they are relationships which are true regardless of the values of $θ$ . This is in contrast to conditional equations such as $sin ⁡ open parenthesis θ close parenthesis equals 1$ which are true for only some values of $θ$ . We will study identities more extensively in Sections 12.1 and 12.2.

While the Reciprocal and Quotient Identities presented in Theorem 11.7 allow us to always reduce problems involving secant, cosecant, tangent and cotangent to problems involving sine and cosine, it is not always convenient to do so.² It is worth taking the time to memorize the tangent and cotangent values of the common angles summarized below.

Tangent and Cotangent Values of Common Angles

$\begin{matrix} θ (degrees) & θ (radians) & \tan (θ) & \cot (θ) \\ 0^{\circ} & 0 & 0 & undefined \\ 30^{\circ} & \frac{π}{6} & \frac{\sqrt{3}}{3} & \sqrt{3} \\ 45^{\circ} & \frac{π}{4} & 1 & 1 \\ 60^{\circ} & \frac{π}{3} & \sqrt{3} & \frac{\sqrt{3}}{3} \\ 90^{\circ} & \frac{π}{2} & undefined & 0 \end{matrix}$

Coupling Theorem 11.7 with the Reference Angle Theorem, Theorem 11.1, we get the following.

Theorem

Generalized Reference Angle Theorem. The values of the circular functions of an angle, if they exist, are the same, up to a sign, of the corresponding circular functions of its reference angle.

More specifically, if $α$ is the reference angle for $θ$ , then:

$\sin (θ) = \pm \sin (α)$ , $\cos (θ) = \pm \cos (α)$ , $\tan (θ) = \pm \tan (α)$

and

$\sec (θ) = \pm \sec (α)$ , $\csc (θ) = \pm \csc (α)$ , $\cot (θ) = \pm \cot (α)$

where the choice of the ( $\pm$ ) depends on the quadrant in which the terminal side of $θ$ lies.

It is high time for an example.

Example 1

Find the exact value of the following, if it exists:
1. $sec ⁡ open parenthesis 60 degrees close parenthesis$
2. $\csc (\frac{7 π}{4})$
3. $tan ⁡ open parenthesis 225 degrees close parenthesis$
4. $\cot (- \frac{7 π}{6})$
Find all angles which satisfy the given equation.
1. $sec ⁡ open parenthesis θ close parenthesis equals 2$
2. $\csc (θ) = - \sqrt{2}$
3. $\tan (θ) = \sqrt{3}$
4. $\cot (θ) = - 1$ .

1. According to Theorem 11.7, $\sec (60^{\circ}) = \frac{1}{\cos (60^{\circ})}$ . Hence, $\sec (60^{\circ}) = \frac{1}{(1 / 2)} = 2$ .
2. Since $\sin (\frac{7 π}{4}) = - \frac{\sqrt{2}}{2}$ , $\csc (\frac{7 π}{4}) = \frac{1}{\sin (\frac{7 π}{4})} = \frac{1}{- \sqrt{2} / 2} = - \frac{2}{\sqrt{2}} = - \sqrt{2}$ .
3. We have two ways to proceed to determine $tan ⁡ open parenthesis 225 degrees close parenthesis$ . First, we can use Theorem 11.7 and note that $\tan (225^{\circ}) = \frac{\sin (225^{\circ})}{\cos (225^{\circ})}$ . Since $\sin (225^{\circ}) = \cos (225^{\circ}) = - \frac{\sqrt{2}}{2}$ , $tan ⁡ open parenthesis 225 degrees close parenthesis equals 1$ .
  Another way to proceed is to note that $225 degrees$ has a reference angle of $45 degrees$ . Per Theorem 11.8, $\tan (225^{\circ}) = \pm \tan (45^{\circ}) = \pm 1$ . Since $225 degrees$ is a Quadrant III angle, where both the $x$ and $y$ coordinates of points are both negative, and tangent is defined as the ratio of coordinates $\frac{y}{x}$ , we know $tan ⁡ open parenthesis 225 degrees close parenthesis greater than 0$ . Hence, $tan ⁡ open parenthesis 225 degrees close parenthesis equals 1$ .
4. As with the previous example, we have two ways to proceed. Using Theorem 11.7, we have $\cot (- \frac{7 π}{6}) = \frac{\cos (- \frac{7 π}{6})}{\sin (- \frac{7 π}{6})}$ . Since $\cos (- \frac{7 π}{6}) = - \frac{\sqrt{3}}{2}$ and $\sin (- \frac{7 π}{6}) = \frac{1}{2}$ , we get $\cot (- \frac{7 π}{6}) = - \sqrt{3}$ .
  Alternatively, we note $- \frac{7 π}{6}$ is a Quadrant II angle with reference angle $\frac{π}{6}$ . Hence, Theorem 11.8 tells us $\cot (- \frac{7 π}{6}) = \pm \cot (\frac{π}{6}) = \pm \sqrt{3}$ . Since $- \frac{7 π}{6}$ is a Quadrant II angle, where the $x$ and $y$ coordinates have different signs, and cotangent is defined as the ratio of coordinates $\frac{x}{y}$ , we know $\cot (- \frac{7 π}{6}) < 0$ . Hence, $\cot (- \frac{7 π}{6}) = - \sqrt{3}$ .
1. To solve $sec ⁡ open parenthesis θ close parenthesis equals 2$ , we convert to cosines and get $\frac{1}{\cos (θ)} = 2$ or $\cos (θ) = \frac{1}{2}$ . This is the exact same equation we solved in Example 11.2.4, number, so we know the answer is: $θ = \frac{π}{3} + 2 π k$ or $θ = \frac{5 π}{3} + 2 π k$ for integers $k$ .
2. From the table of common values, we see $\tan (\frac{π}{3}) = \sqrt{3}$ . According to Theorem 11.8, we know the solutions to $\tan (θ) = \sqrt{3}$ must, therefore, have a reference angle of $\frac{π}{3}$ .
  To find the quadrants in which our solutions lie, we note that tangent is defined as the ratio $\frac{y}{x}$ of points $(x, y)$ on the Unit Circle. Hence, tangent is positive when $x$ and $y$ have the same sign (i.e., when they are both positive or both negative.) This happens in Quadrants I and III.
  In Quadrant I, we get the solutions: $θ = \frac{π}{3} + 2 π k$ for integers $k$ , and for Quadrant III, we get $θ = \frac{4 π}{3} + 2 π k$ for integers $k$ . While these descriptions of the solutions are correct, they can be combined into one list as $θ = \frac{π}{3} + π k$ for integers $k$ . The latter form of the solution is best understood looking at the geometry of the situation in the diagram below.³
  Figure 11.135
  Figure 11.136
3. From the table of common values, we see that $\frac{π}{4}$ has a cotangent of $1$ , which means the solutions to $\cot (θ) = - 1$ have a reference angle of $\frac{π}{4}$ .
  To find the quadrants in which our solutions lie, we note that $\cot (θ) = \frac{x}{y}$ for a point $(x, y)$ on the Unit Circle where $y not equal to 0$ . If $\cot (θ)$ is negative, then $x$ and $y$ must have different signs (i.e., one positive and one negative.) Hence, our solutions lie in Quadrants II and IV.
  Our Quadrant II solution is $θ = \frac{3 π}{4} + 2 π k$ , and for Quadrant IV, we get $θ = \frac{7 π}{4} + 2 π k$ for integers $k$ . As in the previous problem, we can combine these solutions as: $θ = \frac{3 π}{4} + π k$ for integers $k$ .
  Figure 11.137
  Figure 11.138

A few remarks about Example 11.4.1 are in order. First note that the signs ( $\pm$ ) of secant and cosecant are the same as the signs of cosine and sine, respectively.

On the other hand, since tangent and cotangent are defined in terms of the ratios of coordinates $x$ and $y$ , tangent and cotangent are positive in Quadrants I and III (where both $x$ and $y$ have the same sign) and negative in Quadrants II and IV (where $x$ and $y$ have opposite signs.)

The diagram below on the left summarizes which circular functions are positive in which quadrants.

Figure: Postive Circular Functions — **Figure 11.139** Postive Circular Functions

Figure: The period of and is — **Figure 11.140** The period of $\tan (θ)$ and $\cot (θ)$ is $π$

Also note it is no coincidence that both of our solutions to the equations involving tangent and cotangent in Example 11.4.1 could be simplified to just one list of angles differing by multiples of $π$ .

Indeed, any two angles that are $π$ units apart will not only have the same reference angle, but points on their terminal sides on the Unit Circle will be reflections through the origin, as illustrated above on the right.

It follows that the tangent and cotangent of such angles (if defined) will be the same, which means the period of these function is (at most) $π$ .

Using an argument similar to the one we used to establish the period of sine and cosine in Section 11.3, we note that if $\tan (x + p) = \tan (x)$ for all real numbers $x$ , then, in particular, $\tan (p) = \tan (0 + p) = \tan (0) = 0$ . Hence, $p$ is a multiple of $π$ , and the smallest multiple of $π$ is $π$ itself.

Hence, the period of tangent (and cotangent) is $π$ , and we will see the consequences of this both when solving equations in this section and when graphing these functions in Section 11.5.

As with sine and cosine, the circular functions defined in Definition 11.4 agree with those put forth in Definitions B.1 and B.2 in Section B.2 for acute angles situated in right triangles. The argument is identical to the one given in Section 11.2 and is left to the reader.

Moreover, Definition 11.4 can be extended to circles of arbitrary radius $r greater than 0$ using the same similarity arguments in Section 11.2.1 to generalize Definition 11.2 to Theorem 11.3 as summarized below.

Theorem

Suppose $Q (x, y)$ is the point on the terminal side of an angle $θ$ (plotted in standard position) which lies on the circle of radius $r$ , $x^{2} + y^{2} = r^{2}$ . Then:

$\sin (θ) = \frac{y}{r} = \frac{y}{\sqrt{x^{2} + y^{2}}}$
$\cos (θ) = \frac{x}{r} = \frac{x}{\sqrt{x^{2} + y^{2}}}$
$\tan (θ) = \frac{y}{x}$ , provided $x not equal to 0$ .
$\sec (θ) = \frac{r}{x} = \frac{\sqrt{x^{2} + y^{2}}}{x}$ , provided $x not equal to 0$ .
$\csc (θ) = \frac{r}{y} = \frac{\sqrt{x^{2} + y^{2}}}{y}$ , provided $y not equal to 0$ .
$\cot (θ) = \frac{x}{y}$ , provided $y not equal to 0$ .

We make good use of Theorem 11.9 in the following example.

Example 2

Use Theorem 11.9 to solve the following.

Suppose the terminal side of $θ$ , when plotted in standard position, contains the point $Q ⁢ open parenthesis 3, 4 close parenthesis$ . Find the values of the six circular functions of $θ$ .
Suppose $θ$ is a Quadrant IV angle with $\cot (θ) = - 4$ . Find the values of the five remaining circular functions of $θ$ .
Find $\sin (θ)$ , where $\sec (θ) = - \sqrt{5}$ and $θ$ is a Quadrant II angle.
Find $\cos (θ)$ , where $tan ⁡ open parenthesis θ close parenthesis equals 3$ and $π < θ < \frac{3 π}{2}$ .

Since $x equals 3$ and $y equals 4$ , from $x^{2} + y^{2} = r^{2}$ , ${(3)}^{2} + {(4)}^{2} = r^{2}$ so $r squared equals 25$ , or $r equals 5$ . Theorem 11.9 tells us $\sin (θ) = \frac{4}{5}$ , $\cos (θ) = \frac{3}{5}$ , $\tan (θ) = \frac{4}{3}$ , $\sec (θ) = \frac{5}{3}$ , $\csc (θ) = \frac{5}{4}$ , and $\cot (θ) = \frac{3}{4}$ .
In order to use Theorem 11.9, we need to find a point $Q (x, y)$ which lies on the terminal side of $θ$ , when $θ$ is plotted in standard position.
We have that $\cot (θ) = - 4 = \frac{x}{y}$ . Since $θ$ is a Quadrant IV angle, we also know $x greater than 0$ and $y less than 0$ . Rewriting $- 4 = \frac{4}{- 1}$ , we choose⁴ $x equals 4$ and $y = - 1$ so that $r = \sqrt{x^{2} + y^{2}} = \sqrt{{(4)}^{2} + {(- 1)}^{2}} = \sqrt{17}$ .
Applying Theorem 11.9, we find $\sin (θ) = - \frac{1}{\sqrt{17}} = - \frac{\sqrt{17}}{17}$ , $\cos (θ) = \frac{4}{\sqrt{17}} = \frac{4 \sqrt{17}}{17}$ , $\tan (θ) = - \frac{1}{4}$ , $\sec (θ) = \frac{\sqrt{17}}{4}$ , and $\csc (θ) = - \sqrt{17}$ .
Figure 11.141 $Q ⁢ open parenthesis 3, 4 close parenthesis$ lies on a circle of radius $5$ units,
Figure 11.142 $θ$ is Quadrant IV with $\cot (θ) = - 4$ .
To find $\sin (θ)$ using Theorem 11.9, we need to determine the $y$ -coordinate of a point $Q (x, y)$ on the terminal side of $θ$ , when $θ$ is plotted in standard position, and the corresponding radius $r$ .
Since $\sec (θ) = \frac{r}{x}$ and $r greater than 0$ , we rewrite $\sec (θ) = \frac{r}{x} = - \sqrt{5} = \frac{\sqrt{5}}{- 1}$ and take $r = \sqrt{5}$ and $x = - 1$ .
To find $y$ , we substitute $x = - 1$ and $r = \sqrt{5}$ into $x^{2} + y^{2} = r^{2}$ to get ${(- 1)}^{2} + y^{2} = {(\sqrt{5})}^{2}$ . We find $y squared equals 4$ or $y = \pm 2$ . Since $θ$ is a Quadrant II angle, we select $y equals 2$ .
Hence, $\sin (θ) = \frac{y}{r} = \frac{2}{\sqrt{5}} = \frac{2 \sqrt{5}}{5}$ .
We are told $tan ⁡ open parenthesis θ close parenthesis equals 3$ and $π < θ < \frac{3 π}{2}$ , so we know $θ$ is a Quadrant III angle.
To find $\cos (θ)$ using Theorem 11.9, we need to find the $x$ -coordinate of a point $Q (x, y)$ on the terminal side of $θ$ , when $θ$ is plotted in standard position, and the corresponding radius, $r$ .
Since $\tan (θ) = \frac{y}{x}$ and $θ$ is a Quadrant III angle, we rewrite $\tan (θ) = 3 = \frac{- 3}{- 1} = \frac{y}{x}$ and choose $x = - 1$ and $y = - 3$ . From $x^{2} + y^{2} = r^{2}$ , we get $r = \sqrt{10}$ .
Hence, $\cos (θ) = \frac{x}{r} = \frac{- 1}{\sqrt{10}} = - \frac{\sqrt{10}}{10}$ .
Figure 11.143 $θ$ is Quadrant II with $\sec (θ) = - \sqrt{5}$
Figure 11.144 $θ$ is Quadrant III with $tan ⁡ open parenthesis θ close parenthesis equals 3$ .

As we did in Section 11.2.1, we may consider $\tan (t)$ , $\sec (t)$ , $\csc (t)$ , and $\cot (t)$ as functions real numbers by associating each real number $t$ with an angle $θ$ measuring $t$ radians as discussed on page Section 11.2 and using Definition 11.4, or, more generally, Theorem 11.9.

Alternatively, we could define each of these four functions in terms of $f (t) = \sin (t)$ and $g (t) = \cos (t)$ as demonstrated in Theorem 11.7. For example, we could simply define $\sec (t) = \frac{1}{\cos (t)}$ so long as $cos ⁡ open parenthesis t close parenthesis not equal to 0$ .

Either way, we have the means to explore these functions in greater detail. Before doing so, we'll need practice with these additional four circular functions courtesy of the Exercises.

Exercises

In Exercises -, find the exact value or state that it is undefined.

$\tan (\frac{π}{4})$
$\sec (\frac{π}{6})$
$\csc (\frac{5 π}{6})$
$\cot (\frac{4 π}{3})$
$\tan (- \frac{11 π}{6})$
$\sec (- \frac{3 π}{2})$
$\csc (- \frac{π}{3})$
$\cot (\frac{13 π}{2})$
$tan ⁡ open parenthesis 117 ⁢ π close parenthesis$
$\sec (- \frac{5 π}{3})$
$csc ⁡ open parenthesis 3 ⁢ π close parenthesis$
$cot ⁡ open parenthesis minus 5 ⁢ π close parenthesis$
$\tan (\frac{31 π}{2})$
$\sec (\frac{π}{4})$
$\csc (- \frac{7 π}{4})$
$\cot (\frac{7 π}{6})$
$\tan (\frac{2 π}{3})$
$sec ⁡ open parenthesis minus 7 ⁢ π close parenthesis$
$\csc (\frac{π}{2})$
$\cot (\frac{3 π}{4})$

In Exercises -, use the given the information to determine the quadrant in which the terminal side of the angle lies when plotted in standard position.

$sin ⁡ open parenthesis θ close parenthesis greater than 0$ but $tan ⁡ open parenthesis θ close parenthesis less than 0$ .
$cot ⁡ open parenthesis α close parenthesis greater than 0$ but $cos ⁡ open parenthesis α close parenthesis less than 0$ .
$sin ⁡ open parenthesis β close parenthesis greater than 0$ and $tan ⁡ open parenthesis β close parenthesis greater than 0$ .
$cos ⁡ open parenthesis γ close parenthesis greater than 0$ but $cot ⁡ open parenthesis γ close parenthesis less than 0$ .

In Exercises -, use the given the information to find the exact values of the circular functions of $θ$ .

$\sin (θ) = \frac{3}{5}$ with $θ$ in Quadrant II
$\tan (θ) = \frac{12}{5}$ with $θ$ in Quadrant III
$\csc (θ) = \frac{25}{24}$ with $θ$ in Quadrant I
$sec ⁡ open parenthesis θ close parenthesis equals 7$ with $θ$ in Quadrant IV
$\csc (θ) = - \frac{10 \sqrt{91}}{91}$ with $θ$ in Quadrant III
$\cot (θ) = - 23$ with $θ$ in Quadrant II
$\tan (θ) = - 2$ with $θ$ in Quadrant IV.
$\sec (θ) = - 4$ with $θ$ in Quadrant II.
$\cot (θ) = \sqrt{5}$ with $θ$ in Quadrant III.
$\cos (θ) = \frac{1}{3}$ with $θ$ in Quadrant I.
$cot ⁡ open parenthesis θ close parenthesis equals 2$ with $0 < θ < \frac{π}{2}$ .
$csc ⁡ open parenthesis θ close parenthesis equals 5$ with $\frac{π}{2} < θ < π$ .
$\tan (θ) = \sqrt{10}$ with $π < θ < \frac{3 π}{2}$ .
$\sec (θ) = 2 \sqrt{5}$ with $\frac{3 π}{2} < θ < 2 π$ .

In Exercises -, use your calculator to approximate the given value to three decimal places. Make sure your calculator is in the proper angle measurement mode!

$csc ⁡ open parenthesis 78.95 degrees close parenthesis$
$tan ⁡ open parenthesis minus 2.01 close parenthesis$
$cot ⁡ open parenthesis 392.994 close parenthesis$
$sec ⁡ open parenthesis 207 degrees close parenthesis$
$csc ⁡ open parenthesis 5.902 close parenthesis$
$tan ⁡ open parenthesis 39.672 degrees close parenthesis$
$cot ⁡ open parenthesis 3 degrees close parenthesis$
$sec ⁡ open parenthesis 0.45 close parenthesis$

In Exercises -, find all of the angles which satisfy the equation.

$\tan (θ) = \sqrt{3}$
$sec ⁡ open parenthesis θ close parenthesis equals 2$
$\csc (θ) = - 1$
$\cot (θ) = \frac{\sqrt{3}}{3}$
$tan ⁡ open parenthesis θ close parenthesis equals 0$
$sec ⁡ open parenthesis θ close parenthesis equals 1$
$csc ⁡ open parenthesis θ close parenthesis equals 2$
$cot ⁡ open parenthesis θ close parenthesis equals 0$
$\tan (θ) = - 1$
$sec ⁡ open parenthesis θ close parenthesis equals 0$
$\csc (θ) = - \frac{1}{2}$
$\sec (θ) = - 1$
$\tan (θ) = - \sqrt{3}$
$\csc (θ) = - 2$
$\cot (θ) = - 1$

In Exercises -, solve the equation for $t$ . Give exact values.

$cot ⁡ open parenthesis t close parenthesis equals 1$
$\tan (t) = \frac{\sqrt{3}}{3}$
$\sec (t) = - \frac{2 \sqrt{3}}{3}$
$csc ⁡ open parenthesis t close parenthesis equals 0$
$\cot (t) = - \sqrt{3}$
$\tan (t) = - \frac{\sqrt{3}}{3}$
$\sec (t) = \frac{2 \sqrt{3}}{3}$
$\csc (t) = \frac{2 \sqrt{3}}{3}$

In Exercises -, write the given function as a nontrivial decomposition of functions as directed.

For $f (t) = 3 t^{2} + 2 \tan (3 t)$ , find functions $g$ and $h$ so that $f = g + h$ .
For $f (θ) = \sec (θ) - \tan (θ)$ , find functions $g$ and $h$ so that $f = g - h$ .
For $f (t) = - \csc (t) \cot (t)$ , find functions $g$ and $h$ so that $f = g h$ .
For $r (t) = \frac{\tan (3 t)}{t}$ , find functions $f$ and $g$ so $r = \frac{f}{g}$ .
For $T ⁢ open parenthesis θ close parenthesis equals tan ⁡ open parenthesis 4 ⁢ θ close parenthesis$ , find functions $f$ and $g$ so $T equals g ∘ f$ .
For $s (θ) = \sec^{2} (θ)$ , find functions $f$ and $g$ so $s equals g ∘ f$ .
For $L (x) = \ln (\sin (x))$ , find functions $f$ and $g$ so $L equals g ∘ f$ .
For $ℓ (θ) = \ln | \sec (θ) - \tan (θ) |$ , find find functions $f$ , $g$ , and $h$ so $ℓ = h \circ (f - g)$ .
Let $S (t) = \sin (t)$ and $C (t) = \cos (t)$ , $F (t) = \tan (t)$ , and $G (t) = \cot (t)$ . Explain why $F = \frac{S}{C}$ but $F \neq \frac{1}{G}$ .
HINT: Think about domains …
For each function $T (t)$ listed below, compute the average rate of change over the indicated interval.⁵ What trends do you notice? Compare your answer with what you discovered in Section 11.2 number. Be sure your calculator is in radian mode!
$\begin{matrix} T (t) & [- 0.1, 0.1] & [- 0.01, 0.01] & [- 0.001, 0.001] \\ \tan (t) \\ \tan (2 t) \\ \tan (3 t) \\ \tan (4 t) \end{matrix}$
We wish to establish the inequality $\cos (θ) < \frac{\sin (θ)}{θ} < 1$ for $0 < θ < \frac{π}{2} .$ Use the diagram from the beginning of the section, partially reproduced below, to answer the following.
Figure 11.145
1. Show that triangle $O P B$ has area $\frac{1}{2} \sin (θ)$ and triangle $O Q B$ has area $\frac{1}{2} \tan (θ)$ .
2. Show that the circular sector $O P B$ with central angle $θ$ has area $\frac{1}{2} θ$ .
3. Comparing areas, show that $\sin (θ) < θ < \tan (θ)$ for $0 < θ < \frac{π}{2} .$
4. Use the inequality $\sin (θ) < θ$ to show that $\frac{\sin (θ)}{θ} < 1$ for $0 < θ < \frac{π}{2} .$
5. Use the inequality $θ < \tan (θ)$ to show that $\cos (θ) < \frac{\sin (θ)}{θ}$ for $0 < θ < \frac{π}{2} .$ Combine this with the previous part to complete the proof.
Show that $\cos (θ) < \frac{\sin (θ)}{θ} < 1$ also holds for $- \frac{π}{2} < θ < 0$ .
Use the results from Exercises and along with the Squeeze Theorem,⁶ to prove $lim θ → 0$ $\frac{\sin (θ)}{θ} = 1$ .

Answers

$\tan (\frac{π}{4}) = 1$
$\sec (\frac{π}{6}) = \frac{2 \sqrt{3}}{3}$
$\csc (\frac{5 π}{6}) = 2$
$\cot (\frac{4 π}{3}) = \frac{\sqrt{3}}{3}$
$\tan (- \frac{11 π}{6}) = \frac{\sqrt{3}}{3}$
$\sec (- \frac{3 π}{2})$ is undefined
$\csc (- \frac{π}{3}) = - \frac{2 \sqrt{3}}{3}$
$\cot (\frac{13 π}{2}) = 0$
$tan ⁡ open parenthesis 117 ⁢ π close parenthesis equals 0$
$\sec (- \frac{5 π}{3}) = 2$
$csc ⁡ open parenthesis 3 ⁢ π close parenthesis$ is undefined
$cot ⁡ open parenthesis minus 5 ⁢ π close parenthesis$ is undefined
$\tan (\frac{31 π}{2})$ is undefined
$\sec (\frac{π}{4}) = \sqrt{2}$
$\csc (- \frac{7 π}{4}) = \sqrt{2}$
$\cot (\frac{7 π}{6}) = \sqrt{3}$
$\tan (\frac{2 π}{3}) = - \sqrt{3}$
$\sec (- 7 π) = - 1$
$\csc (\frac{π}{2}) = 1$
$\cot (\frac{3 π}{4}) = - 1$
Quadrant II.
Quadrant III.
Quadrant I.
Quadrant IV.
$\sin (θ) = \frac{3}{5}, \cos (θ) = - \frac{4}{5}, \tan (θ) = - \frac{3}{4}, \csc (θ) = \frac{5}{3}, \sec (θ) = - \frac{5}{4}, \cot (θ) = - \frac{4}{3}$
$\sin (θ) = - \frac{12}{13}, \cos (θ) = - \frac{5}{13}, \tan (θ) = \frac{12}{5}, \csc (θ) = - \frac{13}{12}, \sec (θ) = - \frac{13}{5}, \cot (θ) = \frac{5}{12}$
$\sin (θ) = \frac{24}{25}, \cos (θ) = \frac{7}{25}, \tan (θ) = \frac{24}{7}, \csc (θ) = \frac{25}{24}, \sec (θ) = \frac{25}{7}, \cot (θ) = \frac{7}{24}$
$\sin (θ) = \frac{- 4 \sqrt{3}}{7}, \cos (θ) = \frac{1}{7}, \tan (θ) = - 4 \sqrt{3}, \csc (θ) = - \frac{7 \sqrt{3}}{12}, \sec (θ) = 7, \cot (θ) = - \frac{\sqrt{3}}{12}$
$\sin (θ) = - \frac{\sqrt{91}}{10}, \cos (θ) = - \frac{3}{10}, \tan (θ) = \frac{\sqrt{91}}{3}, \csc (θ) = - \frac{10 \sqrt{91}}{91}, \sec (θ) = - \frac{10}{3}, \cot (θ) = \frac{3 \sqrt{91}}{91}$
$\sin (θ) = \frac{\sqrt{530}}{530}, \cos (θ) = - \frac{23 \sqrt{530}}{530}, \tan (θ) = - \frac{1}{23}, \csc (θ) = \sqrt{530}, \sec (θ) = - \frac{\sqrt{530}}{23}, \cot (θ) = - 23$
$\sin (θ) = - \frac{2 \sqrt{5}}{5}, \cos (θ) = \frac{\sqrt{5}}{5}, \tan (θ) = - 2, \csc (θ) = - \frac{\sqrt{5}}{2}, \sec (θ) = \sqrt{5}, \cot (θ) = - \frac{1}{2}$
$\sin (θ) = \frac{\sqrt{15}}{4}, \cos (θ) = - \frac{1}{4}, \tan (θ) = - \sqrt{15}, \csc (θ) = \frac{4 \sqrt{15}}{15}, \sec (θ) = - 4, \cot (θ) = - \frac{\sqrt{15}}{15}$
$\sin (θ) = - \frac{\sqrt{6}}{6}, \cos (θ) = - \frac{\sqrt{30}}{6}, \tan (θ) = \frac{\sqrt{5}}{5}, \csc (θ) = - \sqrt{6}, \sec (θ) = - \frac{\sqrt{30}}{5}, \cot (θ) = \sqrt{5}$
$\sin (θ) = \frac{2 \sqrt{2}}{3}, \cos (θ) = \frac{1}{3}, \tan (θ) = 2 \sqrt{2}, \csc (θ) = \frac{3 \sqrt{2}}{4}, \sec (θ) = 3, \cot (θ) = \frac{\sqrt{2}}{4}$
$\sin (θ) = \frac{\sqrt{5}}{5}, \cos (θ) = \frac{2 \sqrt{5}}{5}, \tan (θ) = \frac{1}{2}, \csc (θ) = \sqrt{5}, \sec (θ) = \frac{\sqrt{5}}{2}, \cot (θ) = 2$
$\sin (θ) = \frac{1}{5}, \cos (θ) = - \frac{2 \sqrt{6}}{5}, \tan (θ) = - \frac{\sqrt{6}}{12}, \csc (θ) = 5, \sec (θ) = - \frac{5 \sqrt{6}}{12}, \cot (θ) = - 2 \sqrt{6}$
$\sin (θ) = - \frac{\sqrt{110}}{11}, \cos (θ) = - \frac{\sqrt{11}}{11}, \tan (θ) = \sqrt{10}, \csc (θ) = - \frac{\sqrt{110}}{10}, \sec (θ) = - \sqrt{11}, \cot (θ) = \frac{\sqrt{10}}{10}$
$\sin (θ) = - \frac{\sqrt{95}}{10}, \cos (θ) = \frac{\sqrt{5}}{10}, \tan (θ) = - \sqrt{19}, \csc (θ) = - \frac{2 \sqrt{95}}{19}, \sec (θ) = 2 \sqrt{5}, \cot (θ) = - \frac{\sqrt{19}}{19}$
$csc ⁡ open parenthesis 78.95 degrees close parenthesis approximately equals 1.019$
$\tan (- 2.01) \approx 2.129$
$cot ⁡ open parenthesis 392.994 close parenthesis approximately equals 3.292$
$\sec (207^{\circ}) \approx - 1.122$
$\csc (5.902) \approx - 2.688$
$tan ⁡ open parenthesis 39.672 degrees close parenthesis approximately equals 0.829$
$cot ⁡ open parenthesis 3 degrees close parenthesis approximately equals 19.081$
$sec ⁡ open parenthesis 0.45 close parenthesis approximately equals 1.111$
$\tan (θ) = \sqrt{3}$ when $θ = \frac{π}{3} + π k$ for any integer $k$
$sec ⁡ open parenthesis θ close parenthesis equals 2$ when $θ = \frac{π}{3} + 2 π k$ or $θ = \frac{5 π}{3} + 2 π k$ for any integer $k$
$\csc (θ) = - 1$ when $θ = \frac{3 π}{2} + 2 π k$ for any integer $k$ .
$\cot (θ) = \frac{\sqrt{3}}{3}$ when $θ = \frac{π}{3} + π k$ for any integer $k$
$tan ⁡ open parenthesis θ close parenthesis equals 0$ when $θ = π k$ for any integer $k$
$sec ⁡ open parenthesis θ close parenthesis equals 1$ when $θ equals 2 ⁢ π ⁢ k$ for any integer $k$
$csc ⁡ open parenthesis θ close parenthesis equals 2$ when $θ = \frac{π}{6} + 2 π k$ or $θ = \frac{5 π}{6} + 2 π k$ for any integer $k$ .
$cot ⁡ open parenthesis θ close parenthesis equals 0$ when $θ = \frac{π}{2} + π k$ for any integer $k$
$\tan (θ) = - 1$ when $θ = \frac{3 π}{4} + π k$ for any integer $k$
$sec ⁡ open parenthesis θ close parenthesis equals 0$ never happens
$\csc (θ) = - \frac{1}{2}$ never happens
$\sec (θ) = - 1$ when $θ = π + 2 π k = (2 k + 1) π$ for any integer $k$
$\tan (θ) = - \sqrt{3}$ when $θ = \frac{2 π}{3} + π k$ for any integer $k$
$\csc (θ) = - 2$ when $θ = \frac{7 π}{6} + 2 π k$ or $θ = \frac{11 π}{6} + 2 π k$ for any integer $k$
$\cot (θ) = - 1$ when $θ = \frac{3 π}{4} + π k$ for any integer $k$
$cot ⁡ open parenthesis t close parenthesis equals 1$ when $t = \frac{π}{4} + π k$ for any integer $k$
$\tan (t) = \frac{\sqrt{3}}{3}$ when $t = \frac{π}{6} + π k$ for any integer $k$
$\sec (t) = - \frac{2 \sqrt{3}}{3}$ when $t = \frac{5 π}{6} + 2 π k$ or $t = \frac{7 π}{6} + 2 π k$ for any integer $k$
$csc ⁡ open parenthesis t close parenthesis equals 0$ never happens
$\cot (t) = - \sqrt{3}$ when $t = \frac{5 π}{6} + π k$ for any integer $k$
$\tan (t) = - \frac{\sqrt{3}}{3}$ when $t = \frac{5 π}{6} + π k$ for any integer $k$
$\sec (t) = \frac{2 \sqrt{3}}{3}$ when $t = \frac{π}{6} + 2 π k$ or $t = \frac{11 π}{6} + 2 π k$ for any integer $k$
$\csc (t) = \frac{2 \sqrt{3}}{3}$ when $t = \frac{π}{3} + 2 π k$ or $t = \frac{2 π}{3} + 2 π k$ for any integer $k$
One solution is $g ⁢ open parenthesis t close parenthesis equals 3 ⁢ t squared$ and $h ⁢ open parenthesis t close parenthesis equals 2 ⁢ tan ⁡ open parenthesis 3 ⁢ t close parenthesis$ .
One solution is $g (θ) = \sec (θ)$ and $h (θ) = \tan (θ)$ .
One solution is $g (t) = - \csc (t)$ and $h (t) = \cot (t)$ .
One solution is $f ⁢ open parenthesis t close parenthesis equals tan ⁡ open parenthesis 3 ⁢ t close parenthesis$ and $g (t) = t$ .
One solution is $f ⁢ open parenthesis θ close parenthesis equals 4 ⁢ θ$ and $g (θ) = \tan (θ)$ .
Since $\sec^{2} (θ) = {(\sec (θ))}^{2}$ , one solution is $f (θ) = \sec (θ)$ and $g (θ) = θ^{2}$ .
One solution is $f (x) = \sin (x)$ and $g (x) = \ln (x)$ .
One solution is $f (θ) = \sec (θ)$ , $g (θ) = \tan (θ)$ , and $h (θ) = \ln | θ |$ .
As we zoom in towards $0$ , the average rate of change of $\tan (k t)$ approaches $k$ . This is the same trend we observed for $\sin (k t)$ in Section 11.2 number.
$\begin{matrix} T (t) & [- 0.1, 0.1] & [- 0.01, 0.01] & [- 0.001, 0.001] \\ \tan (t) & \approx 1.0033 & \approx 1 & \approx 1 \\ \tan (2 t) & \approx 2.0271 & \approx 2.0003 & \approx 2 \\ \tan (3 t) & \approx 3.0933 & \approx 3.0009 & \approx 3 \\ \tan (4 t) & \approx 4.2279 & \approx 4.0021 & \approx 4 \end{matrix}$

Compare this with the definition given in Section 1.2.4. ↩
As we shall see shortly, when solving equations involving secant and cosecant, we usually convert back to cosines and sines. However, when solving for tangent or cotangent, we usually stick with what we're dealt. ↩
See Example 11.2.4 number in Section 11.2 for another example of this kind of simplification of the solution. ↩
We could have just as easily chosen $x equals 8$ and $y = - 2$ - just so long as $x greater than 0$ , $y less than 0$ and $\frac{x}{y} = - 4$ . ↩
See Definition 1.8 in Section 1.2.4 for a review of this concept, as needed. ↩
Theorem 10.2 ↩

Adapted from Precalculus, Preliminary 4th Edition (integrated calculus), by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.