Reciprocal Trigonometric Ratios and Pythagorean Identities
A floodlight on the floor of a school courtyard lights up a wall. If the beam makes an angle $\alpha$ with the ground and you know the distance to the wall, how do you find the length of the beam without dividing every time? That is what three new ratios are for: the secant, the cosecant, and the cotangent. They are just the flipped fractions of the cosine, the sine, and the tangent, but they save many steps when simplifying, verifying identities, and solving equations.
Three new ratios: secant, cosecant, and cotangent
Each of the three new ratios is, by definition, the multiplicative inverse of one of the ratios you already know.
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 $\sec \alpha$.
Values at special angles
Since they are direct reciprocals, the values of secant, cosecant, and cotangent come from inverting the table of sine, cosine, and tangent that you already know.
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
The fundamental identity $\operatorname{sen}^2\alpha+\cos^2\alpha=1$ has two versions with reciprocals. They are obtained by dividing the whole equation by $\cos^2\alpha$ or by $\operatorname{sen}^2\alpha$.
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 $\operatorname{sen}\alpha$; $\cot\alpha$, the same as $\tan\alpha$. The table of signs you already know works for the reciprocals.
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 and verifying identities
To simplify expressions with reciprocals, rewrite everything in terms of sine and cosine: this is usually the shortest path to cancel factors. Verifying an identity means transforming one side until it becomes equal to the other.
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: $\frac{\sec x}{\csc x} \times \cot^2 x$.
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: $\frac{\tan x + \sec x - 1}{\tan x - \sec x + 1} = \tan x + \sec x$.
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: $\frac{\sec^4 x - 1}{\tan^2 x} - \tan^2 x$.
Prove the following trigonometric identity, showing the necessary algebraic steps:…
Prove the following trigonometric identity, showing the necessary algebraic steps: $\frac{\csc \alpha + \cot \alpha}{\sec \alpha + \tan \alpha} = \frac{\sec \alpha - \tan \alpha}{\csc \alpha - \cot \alpha}$.
Equations and problems with reciprocals
To solve an equation with reciprocals, solve for the ratio and convert to sine, cosine, or tangent; then look for the angles in the requested interval. If two different ratios appear, use a Pythagorean identity to leave just one.
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 $\cot \alpha = \frac{3}{4}$.
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 $[0^\circ, 360^\circ)$.
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 $\sin \theta_i = n \times \sin \theta_r$).
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 $x, y \in [0, \pi]$:
$$ \begin{cases} \csc x + \sec y = 3 \\ \csc x - \sec y = 1 \end{cases} $$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 $\cot \theta = \sqrt[3]{\frac{a}{b}}$.
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 $\csc x + \cot x = \sqrt{3}$.
Back to the beginning: the floodlight's beam measures $L=10\sec\alpha$: the secant turns the horizontal distance into the length of the hypotenuse in a single step, without dividing by the cosine every time. In the same way, a ramp that rises a height $h$ with inclination $\alpha$ is the hypotenuse of a right triangle, so its length is $L=\dfrac{h}{\sin\alpha}=h\csc\alpha$. For example, for $h=1$ m and $\alpha=30^\circ$, since $\csc(30^\circ)=2$, the ramp measures $L=2$ m.