Sine, cosine and tangent each have a "partner": their reciprocal. They appear constantly when simplifying expressions and solving trigonometric equations, and also in physics (for example, the length of a ramp that covers a height $h$ at an inclination $\alpha$ is expressed with a cosecant). They are just inverted fractions, but they are worth mastering because they save steps in half the exercises in this block.
Each of the three new ratios is, by definition, the reciprocal of one of the ratios you already know.
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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 $\sec \alpha$.
Since they are direct reciprocals, the exact values of secant, cosecant and cotangent come from inverting the table of sine, cosine and tangent that you already know.
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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.
The fundamental identity $\sin^2\alpha+\cos^2\alpha=1$ has two "reciprocal" versions, which are obtained by dividing the whole equation 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 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 $\sec \alpha = -\frac{13}{5}$ 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).
Since $\sec\alpha$ has the same sign as $\cos\alpha$, and $\csc\alpha$ and $\cot\alpha$ have the same sign as $\sin\alpha$ and $\tan\alpha$ respectively, the table of signs for the three original ratios also works for the reciprocals, without learning anything new.
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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.
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 $E = \frac{\sec x + \tan x}{\cot x}$.
To simplify expressions involving reciprocal ratios, it is best to convert everything to sine and cosine: this is usually the shortest route to getting factors to cancel.
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Simplify the following trigonometric expression as much as possible: $\cos x \cdot \sec x + \sin x \cdot \csc x$.
Simplify the following trigonometric expression as much as possible: $\cos x \cdot \sec x + \sin x \cdot \csc x$.
Back to the beginning: A ramp that covers a height $h$ at inclination $\alpha$ is the hypotenuse of a right-angled triangle, so its length is $L = \frac{h}{\sin\alpha} = h\csc\alpha$. For example, for $h = 1$ m and $\alpha = 30^\circ$, since $\csc(30^\circ) = 2$, the ramp measures $L = 2$ m.
Summary
Let's practice! 10 medium exercises in this section