Reciprocal Trigonometric Ratios
An antenna is installed on the roof of a building in Mexico City, and a taut cable holds it to the ground. If you know the height of the antenna and the angle of the cable, how do you get the length of the cable without dividing by a sine every time? Reciprocal ratios do exactly that: they are inverted fractions that shorten calculations, simplify expressions, and show up in optics and engineering. By the end of this page you will solve that cable problem.
Three new ratios: secant, cosecant, and cotangent
You already know sine, cosine, and tangent. Each one has a "partner" that is its multiplicative inverse.
In a right triangle: $\sec\alpha=\dfrac{\text{hypotenuse}}{\text{adjacent leg}}$, $\csc\alpha=\dfrac{\text{hypotenuse}}{\text{opposite leg}}$, and $\cot\alpha=\dfrac{\text{adjacent}}{\text{opposite}}$.
For the first exercise you need $\csc\alpha=\dfrac{1}{\sin\alpha}$ and to get $\cos\alpha$ from the 3-4-5 triangle.
Given that $\sin \alpha = \frac{3}{5}$ and that $\alpha$ is an angle in the first quadrant, calculate the exact value of $\csc \alpha$ and $\sec \alpha$.
Given that $\sin \alpha = \frac{3}{5}$ and that $\alpha$ is an angle in the first quadrant, calculate the exact value of $\csc \alpha$ and $\sec \alpha$.
Special values and reduction to the first quadrant
The special values come from inverting the sine, cosine, and tangent values you already know.
For angles in other quadrants you use the reference angle: you compute with the equivalent acute angle and the quadrant decides the sign.
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.
Calculate the exact value of $\cot \frac{\pi}{6}$ and $\csc \frac{\pi}{4}$.
Calculate the exact value of $\cot \frac{\pi}{6}$ and $\csc \frac{\pi}{4}$.
Evaluate the following expression giving its exact value, without using decimal approximations: $\sec(120^\circ) + \csc(210^\circ) - \cot(315^\circ)$.
Evaluate the following expression giving its exact value, without using decimal approximations: $\sec(120^\circ) + \csc(210^\circ) - \cot(315^\circ)$.
Pythagorean identities with the reciprocals
The fundamental identity $\sin^2\alpha+\cos^2\alpha=1$ has two other versions that are obtained by dividing it by $\cos^2\alpha$ or by $\sin^2\alpha$.
If $\tan \beta = \frac{4}{3}$ and $0 < \beta < \frac{\pi}{2}$, use the fundamental identity $1 + \tan^2 \beta = \sec^2 \beta$ to calculate $\sec \beta$.
If $\tan \beta = \frac{4}{3}$ and $0 < \beta < \frac{\pi}{2}$, use the fundamental identity $1 + \tan^2 \beta = \sec^2 \beta$ to calculate $\sec \beta$.
Given that $\cos \theta = \frac{1}{3}$, calculate the exact numerical value of the expression $\sec^2 \theta - 1$.
Given that $\cos \theta = \frac{1}{3}$, calculate the exact numerical value of the expression $\sec^2 \theta - 1$.
Given that $\sec \alpha = -\frac{13}{5}$ and that $\pi < \alpha < \frac{3\pi}{2}$, calculate the exact value of the remaining trigonometric ratios (sine, cosine,…
Given that $\sec \alpha = -\frac{13}{5}$ and that $\pi < \alpha < \frac{3\pi}{2}$, calculate the exact value of the remaining trigonometric ratios (sine, cosine, tangent, cosecant and cotangent).
Sign and existence by quadrant
$\sec\alpha$ has the same sign as $\cos\alpha$; $\csc\alpha$, the same as $\sin\alpha$; and $\cot\alpha$, the same as $\tan\alpha$. The table of signs you already know works the same way.
Compare sine with cosecant on the graph: where sine is $0$, cosecant does not exist, and where sine is $1$ or $-1$, they coincide.
An angle $\gamma$ satisfies $\sec \gamma < 0$ and $\csc \gamma > 0$. Determine, with reasons, which quadrant of the unit circle the angle $\gamma$ lies in.
An angle $\gamma$ satisfies $\sec \gamma < 0$ and $\csc \gamma > 0$. Determine, with reasons, which quadrant of the unit circle the angle $\gamma$ lies in.
If $\cot \alpha = 2$ and $\alpha$ belongs to the third quadrant, determine the sign and exact value of $\tan \alpha$ and $\csc \alpha$.
If $\cot \alpha = 2$ and $\alpha$ belongs to the third quadrant, determine the sign and exact value of $\tan \alpha$ and $\csc \alpha$.
Given that $\csc x = 3$ and that the angle $x$ belongs to the second quadrant, calculate the exact value of the expression $E = \frac{\sec x + \tan x}{\cot x}$.
Given that $\csc x = 3$ and that the angle $x$ belongs to the second quadrant, calculate the exact value of the expression $E = \frac{\sec x + \tan x}{\cot x}$.
Simplifying expressions
The safest strategy is to convert everything to sine and cosine: that way factors cancel. Afterward, if it helps, you can write the result again with a reciprocal.
Simplify the following trigonometric expression as much as possible: $\cos x \times \sec x + \sin x \times \csc x$.
Simplify the following trigonometric expression as much as possible: $\cos x \times \sec x + \sin x \times \csc x$.
Simplify the following trigonometric expression to obtain a single ratio: $\frac{\sec x}{\csc x} \times \cot^2 x$.
Simplify the following trigonometric expression to obtain a single ratio: $\frac{\sec x}{\csc x} \times \cot^2 x$.
Simplify the following trigonometric expression and express it in terms of a single trigonometric ratio or numerical constant: $\frac{\sec^4 x - 1}{\tan^2 x} - \tan^2 x$.
Simplify the following trigonometric expression and express it in terms of a single trigonometric ratio or numerical constant: $\frac{\sec^4 x - 1}{\tan^2 x} - \tan^2 x$.
Proving identities
Proving an identity means showing that two expressions are equal for every angle where they are defined. The rules of the game:
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Work on one side only (usually the more complicated one) until you reach the other. Don't move terms from one side to the other.
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Convert everything to sine and cosine, or use a Pythagorean identity.
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If there is a sum or difference in the denominator, multiply by its conjugate.
Prove the following trigonometric identity: $\frac{1}{\sec x + \tan x} = \sec x - \tan x$.
Prove the following trigonometric identity: $\frac{1}{\sec x + \tan x} = \sec x - \tan x$.
Prove that for any angle $\alpha$ where the ratios are defined, the following equality holds: $\cot^2 \alpha - \cos^2 \alpha = \cot^2 \alpha \times \cos^2 \alpha$.
Prove that for any angle $\alpha$ where the ratios are defined, the following equality holds: $\cot^2 \alpha - \cos^2 \alpha = \cot^2 \alpha \times \cos^2 \alpha$.
Prove the following trigonometric identity: $\frac{\tan x + \sec x - 1}{\tan x - \sec x + 1} = \tan x + \sec x$.
Prove the following trigonometric identity: $\frac{\tan x + \sec x - 1}{\tan x - \sec x + 1} = \tan x + \sec x$.
A trick: $\csc\alpha\pm\cot\alpha$
Just as $\sec^2\alpha-\tan^2\alpha=1$, from $1+\cot^2\alpha=\csc^2\alpha$ you get $\csc^2\alpha-\cot^2\alpha=1$.
This is useful for proving identities with quotients and also for solving equations: if you know $\csc x+\cot x=k$, then $\csc x-\cot x=\dfrac1k$, and by adding and subtracting you get each ratio separately.
Prove the following trigonometric identity, showing the necessary algebraic steps:…
Prove the following trigonometric identity, showing the necessary algebraic steps: $\frac{\csc \alpha + \cot \alpha}{\sec \alpha + \tan \alpha} = \frac{\sec \alpha - \tan \alpha}{\csc \alpha - \cot \alpha}$.
Determine all values of $x \in [0, 2\pi)$ satisfying the trigonometric equation $\csc x + \cot x = \sqrt{3}$.
Determine all values of $x \in [0, 2\pi)$ satisfying the trigonometric equation $\csc x + \cot x = \sqrt{3}$.
Equations with reciprocals
To solve an equation: (1) solve for the ratio, (2) convert it to sine, cosine, or tangent if needed, (3) find all the angles in the interval using the reference angle and the quadrant.
Solve the following trigonometric equation, stating all solutions in the interval $[0, 2\pi]$: $2\csc x + 4 = 0$.
Solve the following trigonometric equation, stating all solutions in the interval $[0, 2\pi]$: $2\csc x + 4 = 0$.
Solve the trigonometric equation $\sec^2 x + \tan x = 3$ in the interval $[0^\circ, 360^\circ)$.
Solve the trigonometric equation $\sec^2 x + \tan x = 3$ in the interval $[0^\circ, 360^\circ)$.
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 $x, y \in [0, \pi]$:
$$ \begin{cases} \csc x + \sec y = 3 \\ \csc x - \sec y = 1 \end{cases} $$Applications: cables, rays, and beams
When the known quantity is one leg and you are looking for the hypotenuse, secant and cosecant give you the length with a single multiplication: $L=\text{leg}\cdot\sec\alpha$ if the leg is adjacent, and $L=\text{leg}\cdot\csc\alpha$ if the leg is opposite.
Back to the beginning: the antenna cable measures "height$\cdot\csc\alpha$" or "distance$\cdot\sec\alpha$," depending on which leg you know. With a single product you get its length, without dividing by a sine or a cosine.
A spotlight illuminates a wall from the ground. The horizontal distance from the spotlight to the base of the wall is $10$ m. The upper light ray forms an angle $\alpha$…
A spotlight illuminates a wall from the ground. The horizontal distance from the spotlight to the base of the wall is $10$ m. 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 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 length $L = h \times \sec \theta_r$, where $h = 5$ cm is the thickness of the block and $\theta_r$ is the angle of refraction. If the ray is incident from air at an angle $\theta_i = 60^\circ$ and the refractive index of the glass is $n = \sqrt{2}$, calculate the exact length $L$ traveled by the laser inside the block. (Note: Snell's law states that $\sin \theta_i = n \times \sin \theta_r$).
An L-shaped hallway has two sections of widths $a = 2$ m and $b = 3$ m that intersect at a right angle. A rigid beam of length $L$ gets stuck while attempting to turn…
An L-shaped hallway has two sections of widths $a = 2$ m and $b = 3$ m that intersect at a right angle. A rigid beam of length $L$ gets stuck while attempting 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 makes with one of the outer walls can be modeled as $L(\theta) = a \sec \theta + b \csc \theta$. Calculate the exact value of $L(\theta)$ when $\cot \theta = \sqrt[3]{\frac{a}{b}}$.