Secant, Cosecant, Cotangent - SAT Math

Card 0 of 48

Question

Find the value of the trigonometric function in fraction form for triangle .

What is the secant of $\angle A$ ?

Answer

The value of the secant of an angle is the value of the hypotenuse over the adjacent.

Therefore:

$sec \angle A = \frac{hypotenuse}{adjacent} = \frac{25}{24}$

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Question

Which of the following is the equivalent to $\frac{1}{\csc\theta}$ ?

Answer

Since $\csc\theta=\frac{1}{\sin\theta}$ :

$\frac{1}{\csc\theta}=\frac{1}{\frac{1}{\sin\theta}}=1*\frac{\sin\theta}{1}=\sin\theta$

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Question

If $\csc \theta < 0$ and $\tan \theta > 0$ , then which of the following must be true about $\theta$ .

Answer

Since cosecant is negative, theta must be in quadrant III or IV.

Since tangent is positive, it must be in quadrant I or III.

Therefore, theta must be in quadrant III.

Using a unit circle we can see that quadrant III is when theta is between $\pi$ and $\frac{3\pi}{2}$ .

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Question

If $\cot \alpha = 5$ and $\sin \alpha < 0$ , what is the value of $\sec \alpha$ ?

Answer

Since cotangent is positive and sine is negative, alpha must be in quadrant III. $\cot \alpha = \frac{x}{y}$ then implies that is a point on the terminal side of alpha.

$r=\sqrt{(x^2+y^2)}=\sqrt{26}$

$\sec \alpha = \frac{r}{x} = \frac{\sqrt{26}}{-5}$

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Question

For the above triangle, what is $\sec (\theta)$ if , and ?

Answer

Secant is the reciprocal of cosine.

$\sec \left ( \theta\right ) = \frac{1}{\cos \left ( \theta\right )}$

It's formula is:

$\sec(\theta) = \frac{\textup{hypotenuse}}{\textup{adjacent}}$

Substituting the values from the problem we get,

$\sec(\theta) = \frac{17}{15} = 1.13$

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Question

For the above triangle, what is $\cot (\theta)$ if , and ?

Answer

Cotangent is the reciprocal of tangent.

$\cot \left ( \theta\right ) = \frac{1}{\tan \left ( \theta\right )}$

It's formula is:

$\cot(\theta) = \frac{\textup{adjacent}}{\textup{opposite}}$

Substituting the values from the problem we get,

$\cot(\theta) = \frac{24}{18} = 1.33$

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Question

The point lies on the terminal side of an angle in standard position. Find the secant of the angle.

Answer

Secant is defined to be the ratio of to where is the distance from the origin.

The Pythagoreanr Triple 5, 12, 13 helps us realize that .

Since , the answer is $\frac{13}{12}$ .

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Question

Given angles and in quadrant I, and given,

$\sin x = \frac{3}{5}$ and $\cos y = \frac {5}{13}$ ,

find the value of $\csc (x+y)$ .

Answer

Use the following trigonometric identity to solve this problem.

$\csc (x+y) = \frac {1}{\sin (x+y)} = \frac {1}{\sin x \cos y + \cos x \sin y}$

Using the Pythagorean triple 3,4,5, it is easy to find $\cos x = \frac {4}{5}$ .

Using the Pythagorean triple 5,12,13, it is easy to find $\sin y = \frac {12}{13}$ .

So substituting all four values into the top equation, we get

$\csc (x+y) = \frac {1}{\frac {3}{5} \cdot \frac {5}{13} + \frac {4}{5} \cdot \frac {12}{13}} = \frac {65}{63}$

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Question

Evaluate:

Answer

Evaluate each term separately.

$sec(60)= \frac{1}{cos(60)}=\frac{1}{0.5}= 2$

$csc(45)= \frac{1}{sin(45)}= \frac{1}{\frac{\sqrt2}{2}}= \frac{2}{\sqrt2}\times\frac{\sqrt2}{\sqrt2}= \frac{2\sqrt2}{2}=\sqrt2$

$sec(60)-csc(45)=2-\frac{2\sqrt2}{2}=2-\sqrt2$

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Question

Determine the value of $cot(\frac{\pi}{4})$ .

Answer

Rewrite $cot(\frac{\pi}{4})$ in terms of sine and cosine.

$cot(\frac{\pi}{4})= \frac{cos(\frac{\pi}{4})}{sin(\frac{\pi}{4})}=\frac{\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}=1$

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Question

Pick the ratio of side lengths that would give sec C.

Answer

$\sec\theta=\frac{1}{\cos\theta }$

Find the ratio of Cosine and take the reciprocal.

$\cos\theta=\frac{AC}{CB}\rightarrow \sec\theta=\frac{1}{\frac{AC}{CB}}=\mathbf{\frac{CB}{AC}}$

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Question

Evaluate:

Answer

Recall that $sec(x) = \frac{1}{cos(x)}$ and $csc(x) = \frac{1}{sin(x)}$ .

Rewrite the expression.

$sec(30)-csc(60) = \frac{1}{cos(30)}-\frac{1}{sin(60)}$

The value of $cos(30) = \frac{\sqrt3}{2}$ and $sin(60) =\frac{ \sqrt{3}}{2}$ .

Since these values are similar, our resulting answer is zero upon substitution.

$\frac{1}{\frac{\sqrt3}{2}}-\frac{1}{\frac{\sqrt3}{2}} = 0$

The answer is:

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Question

Find the value of the trigonometric function in fraction form for triangle .

What is the secant of $\angle A$ ?

Answer

The value of the secant of an angle is the value of the hypotenuse over the adjacent.

Therefore:

$sec \angle A = \frac{hypotenuse}{adjacent} = \frac{25}{24}$

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Question

Which of the following is the equivalent to $\frac{1}{\csc\theta}$ ?

Answer

Since $\csc\theta=\frac{1}{\sin\theta}$ :

$\frac{1}{\csc\theta}=\frac{1}{\frac{1}{\sin\theta}}=1*\frac{\sin\theta}{1}=\sin\theta$

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Question

If $\csc \theta < 0$ and $\tan \theta > 0$ , then which of the following must be true about $\theta$ .

Answer

Since cosecant is negative, theta must be in quadrant III or IV.

Since tangent is positive, it must be in quadrant I or III.

Therefore, theta must be in quadrant III.

Using a unit circle we can see that quadrant III is when theta is between $\pi$ and $\frac{3\pi}{2}$ .

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Question

If $\cot \alpha = 5$ and $\sin \alpha < 0$ , what is the value of $\sec \alpha$ ?

Answer

Since cotangent is positive and sine is negative, alpha must be in quadrant III. $\cot \alpha = \frac{x}{y}$ then implies that is a point on the terminal side of alpha.

$r=\sqrt{(x^2+y^2)}=\sqrt{26}$

$\sec \alpha = \frac{r}{x} = \frac{\sqrt{26}}{-5}$

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Question

For the above triangle, what is $\sec (\theta)$ if , and ?

Answer

Secant is the reciprocal of cosine.

$\sec \left ( \theta\right ) = \frac{1}{\cos \left ( \theta\right )}$

It's formula is:

$\sec(\theta) = \frac{\textup{hypotenuse}}{\textup{adjacent}}$

Substituting the values from the problem we get,

$\sec(\theta) = \frac{17}{15} = 1.13$

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Question

For the above triangle, what is $\cot (\theta)$ if , and ?

Answer

Cotangent is the reciprocal of tangent.

$\cot \left ( \theta\right ) = \frac{1}{\tan \left ( \theta\right )}$

It's formula is:

$\cot(\theta) = \frac{\textup{adjacent}}{\textup{opposite}}$

Substituting the values from the problem we get,

$\cot(\theta) = \frac{24}{18} = 1.33$

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Question

The point lies on the terminal side of an angle in standard position. Find the secant of the angle.

Answer

Secant is defined to be the ratio of to where is the distance from the origin.

The Pythagoreanr Triple 5, 12, 13 helps us realize that .

Since , the answer is $\frac{13}{12}$ .

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Question

Given angles and in quadrant I, and given,

$\sin x = \frac{3}{5}$ and $\cos y = \frac {5}{13}$ ,

find the value of $\csc (x+y)$ .

Answer

Use the following trigonometric identity to solve this problem.

$\csc (x+y) = \frac {1}{\sin (x+y)} = \frac {1}{\sin x \cos y + \cos x \sin y}$

Using the Pythagorean triple 3,4,5, it is easy to find $\cos x = \frac {4}{5}$ .

Using the Pythagorean triple 5,12,13, it is easy to find $\sin y = \frac {12}{13}$ .

So substituting all four values into the top equation, we get

$\csc (x+y) = \frac {1}{\frac {3}{5} \cdot \frac {5}{13} + \frac {4}{5} \cdot \frac {12}{13}} = \frac {65}{63}$

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