# Trig problem

## Homework Statement

Find cscθ given sec θ = -2 sin θ >0

## Homework Equations

I do not know where to begin or what equations to use.

## The Attempt at a Solution

I am assuming there is a typo in the question and that there should be a comma sec θ = -2, sin θ >0.

If so I got csc θ = 2√3 / 3

SteamKing
Staff Emeritus
Homework Helper

## Homework Statement

Find cscθ given sec θ = -2 sin θ >0

## Homework Equations

I do not know where to begin or what equations to use.

## The Attempt at a Solution

I do not know where to begin or what equations to use.
Why not? How is the secant defined? The cosecant? Those would be good places to start.

I am assuming there is a typo in the question and that there should be a comma sec θ = -2, sin θ >0.

If so I got csc θ = 2√3 / 3

SteamKing
Staff Emeritus
Homework Helper
I am assuming there is a typo in the question and that there should be a comma sec θ = -2, sin θ >0.

If so I got csc θ = 2√3 / 3
Looks good.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Find cscθ given sec θ = -2 sin θ >0

## Homework Equations

I do not know where to begin or what equations to use.

## The Attempt at a Solution

I am assuming there is a typo in the question and that there should be a comma sec θ = -2, sin θ >0.

If so I got csc θ = 2√3 / 3

The question as written (no comma) also makes perfectly good sense; it says that ##0 < \sec \theta = - 2 \sin \theta##, and you need to use the definition of ##\sec## to get a solvable equation. You can get ##\sin \theta## and ##\cos \theta##, then compute ##\csc \theta##, which does NOT equal ##2 \sqrt{3}/3 = 2/\sqrt{3}##.

Last edited:
SammyS
SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

Find cscθ given sec θ = -2 sin θ >0
As Ray points out, the problem as stated makes sense and can be solved. If you are currently studying trig equations in your course, then that's probably not a typo. On the other hand, if you are studying the basics of trigonometric functions early in your course, then it very well may be a typo, similar to what you have concluded.

The statement, ##\ \sec θ = -2 \sin θ >0 \,,\ ## is really two (or three) statements rolled into one.
• ##\ \sec θ >0 \ ##
• ##\ -2 \sin θ >0 \,,\ ## so that ##\ \sin θ <0 \ ##
• ##\ \sec θ = -2 \sin θ ##
The first two tell you what quadrant θ is in.

Dr Transport
recast $sec(\theta)$ as $1/cos(\theta)$ and the solution falls out in about 2 lines...