Your calculator has buttons for $\sin$, $\cos$, and $\tan$, but none for $\sec$, $\csc$, or $\cot$. So how would you find $\sec 60^\circ$? These three "extra" ratios are just the old ones flipped upside down. They show up all the time when you simplify expressions, solve equations, and model real situations such as light bending in glass or a beam turning a corner. We'll answer the calculator question at the end of the page.
Secant, cosecant, and cotangent
Each of the three new ratios is the reciprocal of one you already know. In a right triangle, you just flip the fraction.
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Knowing that $\sin \alpha = \frac{3}{5}$ and that $\alpha$ is an angle in the first quadrant, calculate the exact value of $\csc \alpha$ and $\sec \alpha$.
Knowing that $\sin \alpha = \frac{3}{5}$ and that $\alpha$ is an angle in the first quadrant, calculate the exact value of $\csc \alpha$ and $\sec \alpha$.
You don't need to memorize a new table. Take the values of sine, cosine, and tangent from the special triangles and flip them.
Read each ratio from the triangle and flip it. For example, in the $45^\circ$ triangle, $\sec 45^\circ = \dfrac{\text{hyp}}{\text{adj}} = \dfrac{\sqrt2}{1} = \sqrt2$.
For angles outside Quadrant I, find the reference angle, take the value from the table, and attach the correct sign (see the signs section below).
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Determine the exact value of $\sec 60^\circ$ and $\cot 45^\circ$ without using a calculator.
Determine the exact value of $\sec 60^\circ$ and $\cot 45^\circ$ without using a calculator.
You know $\sin^2\alpha + \cos^2\alpha = 1$. Two more identities hide inside it. Look at a right triangle whose adjacent side is $1$. Its opposite side is $\tan\alpha$ and its hypotenuse is $\sec\alpha$.
By the Pythagorean Theorem, $1^2 + \tan^2\alpha = \sec^2\alpha$. A similar triangle with opposite side $1$ gives the cotangent version. Algebraically, divide $\sin^2\alpha + \cos^2\alpha = 1$ by $\cos^2\alpha$ or by $\sin^2\alpha$.
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If $\tan \beta = \frac{4}{3}$ and $0 < \beta < \frac{\pi}{2}$, use the fundamental relationship $1 + \tan^2 \beta = \sec^2 \beta$ to calculate $\sec \beta$.
If $\tan \beta = \frac{4}{3}$ and $0 < \beta < \frac{\pi}{2}$, use the fundamental relationship $1 + \tan^2 \beta = \sec^2 \beta$ to calculate $\sec \beta$.
Knowing that $\sec \alpha = -\frac{13}{5}$ and that $\pi < \alpha < \frac{3\pi}{2}$, calculate the exact value of the other trigonometric ratios (sine, cosine, tangent,…
Knowing that $\sec \alpha = -\frac{13}{5}$ and that $\pi < \alpha < \frac{3\pi}{2}$, calculate the exact value of the other trigonometric ratios (sine, cosine, tangent, cosecant and cotangent).
$\sec\alpha$ has the same sign as $\cos\alpha$. $\csc\alpha$ has the same sign as $\sin\alpha$. $\cot\alpha$ has the same sign as $\tan\alpha$. A reciprocal never changes sign, so the sign chart you already know works for all six ratios. A common memory aid is All Students Take Calculus: All positive in QI, Sine (and csc) in QII, Tangent (and cot) in QIII, Cosine (and sec) in QIV.
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An angle $\gamma$ satisfies $\sec \gamma < 0$ and $\csc \gamma > 0$. Reason in which quadrant of the unit circle the angle $\gamma$ is located.
An angle $\gamma$ satisfies $\sec \gamma < 0$ and $\csc \gamma > 0$. Reason in which quadrant of the unit circle the angle $\gamma$ is located.
Knowing that $\csc x = 3$ and that the angle $x$ belongs to the second quadrant, calculate the exact value of the expression $E = \frac{\sec x + \tan x}{\cot x}$.
Knowing that $\csc x = 3$ and that the angle $x$ belongs to the second quadrant, calculate the exact value of the expression $E = \frac{\sec x + \tan x}{\cot x}$.
Simplify the following trigonometric expression and express it in terms of a single trigonometric ratio or numerical constant: $\frac{\sec^4 x - 1}{\tan^2 x} - \tan^2 x$.
Simplify the following trigonometric expression and express it in terms of a single trigonometric ratio or numerical constant: $\frac{\sec^4 x - 1}{\tan^2 x} - \tan^2 x$.
To prove an identity, show that one side can be changed into the other using only known facts (definitions and Pythagorean identities). Here are the ground rules:
Start with the more complicated side and work toward the other.
Do not add, subtract, or multiply both sides by something. That would assume the thing you are trying to prove.
Convert to sine and cosine, find common denominators, factor, or multiply by a conjugate.
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Prove that for any angle $\alpha$ where the ratios are defined, the following equality holds: $\cot^2 \alpha - \cos^2 \alpha = \cot^2 \alpha \cdot \cos^2 \alpha$.
Prove that for any angle $\alpha$ where the ratios are defined, the following equality holds: $\cot^2 \alpha - \cos^2 \alpha = \cot^2 \alpha \cdot \cos^2 \alpha$.
Two more patterns come up. First, a sum of a reciprocal and a related ratio, such as $\csc x + \cot x = k$. Second, a system in which the unknowns are the values of the ratios themselves.
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Determine all values of $x \in [0, 2\pi)$ that satisfy the trigonometric equation $\csc x + \cot x = \sqrt{3}$.
Determine all values of $x \in [0, 2\pi)$ that satisfy the trigonometric equation $\csc x + \cot x = \sqrt{3}$.
Solve the following system of trigonometric equations for values of $x, y \in [0, \pi]$: $$ \begin{cases} \csc x + \sec y = 3 \\ \csc x - \sec y = 1 \end{cases} $$
Solve the following system of trigonometric equations for values of $x, y \in [0, \pi]$:
$$ \begin{cases} \csc x + \sec y = 3 \\ \csc x - \sec y = 1 \end{cases} $$
Secant and cosecant naturally describe lengths of slanted segments: a cable, a ray of light, a ladder. If the horizontal distance is $d$ and the angle with the ground is $\alpha$, then $\cos\alpha = \dfrac{d}{L}$, so $L = d\sec\alpha$. If the vertical side is $h$, then $L = h\csc\alpha$.
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A spotlight illuminates a wall from the ground. The horizontal distance from the spotlight to the base of the wall is $10$ ft. The upper light ray forms an angle…
A spotlight illuminates a wall from the ground. The horizontal distance from the spotlight to the base of the wall is $10$ ft. The upper light ray forms an angle $\alpha$ with the ground. The length of the light ray from the spotlight to the highest illuminated point on the wall is $L$. Express $L$ using the secant ratio and calculate its exact value if $\cot \alpha = \frac{3}{4}$.
In an optics experiment, a laser beam passes through a glass block with flat, parallel faces. The path of the beam inside the glass has a length…
In an optics experiment, a laser beam passes through a glass block with flat, parallel faces. The path of the beam inside the glass has a length $L = h \cdot \sec \theta_r$, where $h = 5$ in is the thickness of the block and $\theta_r$ is the angle of refraction. If the beam is incident from the air at an angle $\theta_i = 60^\circ$ and the refractive index of the glass is $n = \sqrt{2}$, calculate the exact value of the length $L$ traveled by the laser inside the block. (Fact: Snell's law states that $\sin \theta_i = n \cdot \sin \theta_r$).
An L-shaped corridor has two sections of widths $a = 2$ ft and $b = 3$ ft that intersect at a right angle. A rigid beam of length $L$ gets stuck when trying to turn the…
An L-shaped corridor has two sections of widths $a = 2$ ft and $b = 3$ ft that intersect at a right angle. A rigid beam of length $L$ gets stuck when trying to turn the corner, touching both outer walls and the inner corner. The length of the beam as a function of the angle $\theta$ it forms with one of the outer walls can be modeled as $L(\theta) = a \csc \theta + b \sec \theta$. Calculate the exact value of $L(\theta)$ when $\cot \theta = \sqrt[3]{\frac{a}{b}}$.
Back to the beginning: the calculator question has an easy answer. To find $\sec 60^\circ$, press $1 \div \cos(60)$, and you get $2$. For $\cot$, use $1 \div \tan$, and for $\csc$, use $1 \div \sin$.
Summary
Let's practice! 8 medium exercises in this section