We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

SAT II Math I : Secant, Cosecant, Cotangent

Study concepts, example questions & explanations for SAT II Math I

Create An Account

All SAT II Math I Resources

6 Diagnostic Tests 113 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #11 : Trigonometry

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

Possible Answers:

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

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

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

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

$\sqrt{26}$ $\displaystyle \sqrt{26}$

Correct answer:

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

Explanation:

Since cotangent is positive and sine is negative, alpha must be in quadrant III. $\cot \alpha = \frac{x}{y}$ $\displaystyle \cot \alpha = \frac{x}{y}$ then implies that $\displaystyle (x,y)=(-5,-1)$ is a point on the terminal side of alpha.

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

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

Report an Error

Example Question #12 : Trigonometry

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

Report an Error

Example Question #11 : Trigonometry

The point (12, 5) $\displaystyle (12, 5)$ lies on the terminal side of an angle in standard position. Find the secant of the angle.

Possible Answers:

$\frac{12}{13}$ $\displaystyle \frac{12}{13}$

$\frac{5}{13}$ $\displaystyle \frac{5}{13}$

$\frac{13}{5}$ $\displaystyle \frac{13}{5}$

$\frac{5}{12}$ $\displaystyle \frac{5}{12}$

$\frac{13}{12}$ $\displaystyle \frac{13}{12}$

Correct answer:

$\frac{13}{12}$ $\displaystyle \frac{13}{12}$

Explanation:

Secant is defined to be the ratio of $\displaystyle r$ to $\displaystyle x$ where $\displaystyle r$ is the distance from the origin.

The Pythagoreanr Triple 5, 12, 13 helps us realize that r = 13 $\displaystyle r = 13$.

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

Report an Error

Example Question #13 : Trigonometry

Given angles $\displaystyle x$ and $\displaystyle y$ in quadrant I, and given,

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

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

Possible Answers:

$\frac {65}{64}$ $\displaystyle \frac {65}{64}$

$\frac {99}{65}$ $\displaystyle \frac {99}{65}$

$\frac {65}{63}$ $\displaystyle \frac {65}{63}$

$\frac {65}{64}$ $\displaystyle \frac {65}{64}$

$\frac {63}{65}$ $\displaystyle \frac {63}{65}$

Correct answer:

$\frac {65}{63}$ $\displaystyle \frac {65}{63}$

Explanation:

Use the following trigonometric identity to solve this problem.

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

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

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

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

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

Report an Error

Example Question #1 : Sec, Csc, Ctan

Find the value of the trigonometric function in fraction form for triangle ABC $\displaystyle ABC$.

Triangle

What is the secant of $\angle A$ $\displaystyle \angle A$?

Possible Answers:

24/7 $\displaystyle 24/7$

7/25 $\displaystyle 7/25$

25/7 $\displaystyle 25/7$

24/25 $\displaystyle 24/25$

$\frac{25}{24}$ $\displaystyle \frac{25}{24}$

Correct answer:

$\frac{25}{24}$ $\displaystyle \frac{25}{24}$

Explanation:

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

Therefore:

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

Report an Error

Example Question #1 : Secant, Cosecant, Cotangent

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

Possible Answers:

$\sec\theta$ $\displaystyle \sec\theta$

$\cot\theta$ $\displaystyle \cot\theta$

$\sin\theta$ $\displaystyle \sin\theta$

$\tan\theta\sin\theta$ $\displaystyle \tan\theta\sin\theta$

$\cos\theta$ $\displaystyle \cos\theta$

Correct answer:

$\sin\theta$ $\displaystyle \sin\theta$

Explanation:

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

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

Report an Error

Example Question #21 : Trigonometry

Soh_cah_toa

For the above triangle, what is $\sec (\theta)$ $\displaystyle \sec (\theta)$ if o = 8 $\displaystyle o = 8$, a = 15 $\displaystyle a = 15$ and h = 17 $\displaystyle h = 17$?

Possible Answers:

0.47 $\displaystyle 0.47$

2.13 $\displaystyle 2.13$

{}1.13 $\displaystyle {}1.13$

0.53 $\displaystyle 0.53$

0.88 $\displaystyle 0.88$

Correct answer:

{}1.13 $\displaystyle {}1.13$

Explanation:

Secant is the reciprocal of cosine.

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

It's formula is:

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

Substituting the values from the problem we get,

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

Report an Error

Example Question #22 : Trigonometry

Soh_cah_toa

For the above triangle, what is $\cot (\theta)$ $\displaystyle \cot (\theta)$ if o = 18 $\displaystyle o = 18$, a = 24 $\displaystyle a = 24$ and h = 30 $\displaystyle h = 30$?

Possible Answers:

0.75 $\displaystyle 0.75$

1.33 $\displaystyle 1.33$

1.25 $\displaystyle 1.25$

1.67 $\displaystyle 1.67$

0.60 $\displaystyle 0.60$

Correct answer:

1.33 $\displaystyle 1.33$

Explanation:

Cotangent is the reciprocal of tangent.

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

It's formula is:

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

Substituting the values from the problem we get,

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

Report an Error

Example Question #1 : Secant, Cosecant, Cotangent

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

Possible Answers:

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

$\infty$ $\displaystyle \infty$

$\displaystyle 0$

$\frac{\sqrt2}{2}$ $\displaystyle \frac{\sqrt2}{2}$

$\displaystyle 1$

Correct answer:

$\displaystyle 1$

Explanation:

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

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

Report an Error

Example Question #8 : Secant, Cosecant, Cotangent

Evaluate: $\displaystyle sec(60)-csc(45)$

Possible Answers:

$2\sqrt2+2$ $\displaystyle 2\sqrt2+2$

$2+\sqrt2$ $\displaystyle 2+\sqrt2$

$2-\sqrt2$ $\displaystyle 2-\sqrt2$

$2-\frac{\sqrt2}{2}$ $\displaystyle 2-\frac{\sqrt2}{2}$

$2+\frac{\sqrt2}{2}$ $\displaystyle 2+\frac{\sqrt2}{2}$

Correct answer:

$2-\sqrt2$ $\displaystyle 2-\sqrt2$

Explanation:

Evaluate each term separately.

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

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

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

Report an Error

View SAT Subject Test in Mathematics Level 1 Tutors

Charlotte
Certified Tutor

University of North Carolina at Chapel Hill, Bachelor in Arts, Linguistics. University of North Carolina at Chapel Hill, Mast...

View SAT Subject Test in Mathematics Level 1 Tutors

Tom
Certified Tutor

University of Wisconsin-Madison, Bachelor of Science, Atmospheric Sciences and Meteorology. Embry-Riddle Aeronautical Univers...

View SAT Subject Test in Mathematics Level 1 Tutors

Cielo
Certified Tutor

Rochester Institute of Technology, Bachelor of Engineering, Biochemical Engineering.

All SAT II Math I Resources

6 Diagnostic Tests 113 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

English Tutors in San Francisco-Bay Area, Reading Tutors in Seattle, GMAT Tutors in San Diego, Statistics Tutors in Phoenix, English Tutors in Miami, Physics Tutors in Miami, French Tutors in Philadelphia, GMAT Tutors in Philadelphia, ISEE Tutors in Dallas Fort Worth, Math Tutors in Washington DC

Popular Courses & Classes

SSAT Courses & Classes in Philadelphia, MCAT Courses & Classes in Washington DC, GMAT Courses & Classes in New York City, GRE Courses & Classes in Phoenix, LSAT Courses & Classes in Boston, Spanish Courses & Classes in San Francisco-Bay Area, ISEE Courses & Classes in Los Angeles, GMAT Courses & Classes in Atlanta, SSAT Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Los Angeles

Popular Test Prep

GRE Test Prep in Miami, MCAT Test Prep in San Diego, SSAT Test Prep in Phoenix, MCAT Test Prep in New York City, ISEE Test Prep in Philadelphia, ACT Test Prep in Denver, GMAT Test Prep in Denver, ISEE Test Prep in Denver, GRE Test Prep in Philadelphia, GMAT Test Prep in Atlanta

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

SAT II Math I : Secant, Cosecant, Cotangent

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #11 : Trigonometry

Example Question #12 : Trigonometry

Example Question #11 : Trigonometry

Example Question #13 : Trigonometry

Example Question #1 : Sec, Csc, Ctan

Example Question #1 : Secant, Cosecant, Cotangent

Example Question #21 : Trigonometry

Example Question #22 : Trigonometry

Example Question #1 : Secant, Cosecant, Cotangent

Example Question #8 : Secant, Cosecant, Cotangent

All SAT II Math I Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: