Secant, Cosecant, Cotangent

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SAT Math › Secant, Cosecant, Cotangent

Questions 1 - 10

1

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

Triangle

What is the secant of $\angle A$ ?

$\frac{25}{24}$

Explanation

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

Therefore:

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

2

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

$-\frac{\sqrt{26}}{5}$

$\frac{\sqrt{26}}{5}$

$-\frac{1}{\sqrt{26}}$

$-\frac{5}{\sqrt{26}}$

$\sqrt{26}$

Explanation

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.

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

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

3

Given angles and in quadrant I, and given,

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

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

$\frac {65}{63}$

$\frac {65}{64}$

$\frac {63}{65}$

$\frac {65}{64}$

$\frac {99}{65}$

Explanation

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

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

4

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

$\frac{13}{12}$

$\frac{13}{5}$

$\frac{5}{12}$

$\frac{12}{13}$

$\frac{5}{13}$

Explanation

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 $\frac{13}{12}$ .

5

Evaluate:

$\sqrt{3}$

$\sqrt3-1$

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

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

Explanation

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:

6

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

$\infty$

$\frac{1}{2}$

$\frac{\sqrt2}{2}$

Explanation

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

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

7

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

$\pi < \theta < \frac{3\pi}{2}$

$0 < \theta < \frac{\pi}{2}$

$\frac{\pi}{2} < \theta < \pi$

$\frac{3\pi}{2} < \theta < 2\pi$

$-\frac{\pi}{2} < \theta < 0$

Explanation

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 $\frac{3\pi}{2}$ .

8

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

$\sin\theta$

$\cos\theta$

$\sec\theta$

$\tan\theta\sin\theta$

$\cot\theta$

Explanation

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

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

9

Soh_cah_toa

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

Explanation

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,

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

10

Soh_cah_toa

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

Explanation

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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