Exercises: Reciprocal trigonometric ratios

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 $\sec \alpha$.

Determine the exact value of $\sec 60^\circ$ and $\cot 45^\circ$ without using a calculator.

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 $\sec \beta$.

Simplify the following trigonometric expression as much as possible: $\cos x \times \sec x + \sin x \times \csc x$.

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.

Given that $\cos \theta = \frac{1}{3}$, calculate the exact numerical value of the expression $\sec^2 \theta - 1$.

Calculate the exact value of $\cot \frac{\pi}{6}$ and $\csc \frac{\pi}{4}$.

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 $\sec \alpha = -\frac{13}{5}$ and that $\pi < \alpha < \frac{3\pi}{2}$, calculate the exact value of the remaining trigonometric ratios (sine, cosine, tangent, cosecant and cotangent).

Simplify the following trigonometric expression to obtain a single ratio: $\frac{\sec x}{\csc x} \times \cot^2 x$.

Prove the following trigonometric identity: $\frac{1}{\sec x + \tan x} = \sec x - \tan x$.

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 $[0, 2\pi]$: $2\csc x + 4 = 0$.

Evaluate the following expression giving its exact value, without using decimal approximations: $\sec(120^\circ) + \csc(210^\circ) - \cot(315^\circ)$.

Given that $\csc x = 3$ and that the angle $x$ belongs to the second quadrant, calculate the exact value of the expression $E = \frac{\sec x + \tan x}{\cot x}$.

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: $\frac{\tan x + \sec x - 1}{\tan x - \sec x + 1} = \tan x + \sec x$.

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 $L = h \times \sec \theta_r$, 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 $\sin \theta_i = n \times \sin \theta_r$).

Simplify the following trigonometric expression and express it in terms of a single trigonometric ratio or numerical constant: $\frac{\sec^4 x - 1}{\tan^2 x} - \tan^2 x$.

Solve the following system of trigonometric equations for values of $x, y \in [0, \pi]$:

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 $\cot \theta = \sqrt[3]{\frac{a}{b}}$.

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 $\csc x + \cot x = \sqrt{3}$.