Question

URGENT: MultipleChoice

Answer 1

What is the domain of validity for \[\csc \theta = \frac{ 1 }{ }\]

a) all real numbers
b) all real numbers except odd multiples of \[\frac{ \pi }{ 2}\]
c) all real numbers except even multiples of \[\frac{ \pi }{2 }\]
d) all real numbers except multiples of pi

Answer 2

*QUESTION:
What is the domain validity for \[\csc \theta = \frac{ 1 }{\sin \theta }\]

Answer 3

if \(\sin \theta=0\) then u will have division by zero...this is not allowed

so we must have \(\sin \theta\neq0\)

Answer 4

are you saying the value has to be greater than zero..? I dont understand you

Answer 5

Are you allowed to choose more than one answer?

Answer 6

no, just one answer. I already know a) is incorrect

Answer 7

Yes, some values will make csc(theta) undefined, so the set of all real numbers is not the domain

Answer 8

but look at choices C and D

Answer 9

if you're looking at even multiples of pi/2, then you're looking at the numbers

0*pi/2 = 0

2*pi/2 = pi

4*pi/2 = 2pi

etc

This is what choice C is saying, but choice D is the same thing since the multiples of pi are 0, pi, 2pi, etc...

Answer 10

choice C specifies only EVEN, no ODDS.
choice D says ALL of them.

Answer 11

so I guess Choice C right?

Answer 12

choice D says all real numbers except multiples of pi correct?

Answer 13

yes, multiples of pi, ODDS and EVEN

Answer 14

But that's the same thing that choice C says, choice C is just stating the same thing in a slightly different way

Answer 15

Choice C says except EVEN multiples....

Answer 16

and the even multiples of pi/2 are 0, pi, 2pi, etc

Answer 17

soo.... what is the domain of validity for cscθ=1/sinθ

Answer 18

basically, the numbers 0, pi, 2pi, ... etc will make csc(x) undefined

The problem I'm having is that choices C and D give the same answer (they just word it differently)

Answer 19

so there's either a typo, or they allow multiple answers (probably a typo though)

Answer 20

let's just leave that one out..... can you rather help me with this one? I have to simplify the trigonometric expression:

\[\frac{ 1 }{ 1 +\sin \theta } + \frac{ 1 }{ 1 -\sin \theta }\]

Answer 21

\[csc(\theta)\in \mathbb{R}, \theta\in \mathbb{R}\backslash\{\pi\mathbb{Z}\}\]
An even number is a number of the form \(n=2k\) for some \(k \in \mathbb{Z}\) which implies that \(n\frac{\pi}{2}=k\pi\), which is the equivalent of the above statement.

Answer 22

So, both C and D are correct answers.

Answer 23

yeah let's move on and you can ask your teacher about it

Answer 24

As @jim_thompson5910 had stated.

Answer 25

a) \[2\cos ^{2}\theta \]


b)\[2\sec^{2}\theta\]


c)\[2\cot^{2}\theta\]

Answer 26

\[\frac{ 1 }{ 1 +\sin \theta } + \frac{ 1 }{ 1 -\sin \theta }\]


\[\frac{ 1(1 -\sin \theta) }{ (1 +\sin \theta)(1 -\sin \theta) } + \frac{ 1 }{ 1 -\sin \theta }\]


\[\frac{ 1 -\sin \theta }{ (1 +\sin \theta)(1 -\sin \theta) } + \frac{ 1 }{ 1 -\sin \theta }\]


\[\frac{ 1 -\sin \theta }{ (1 +\sin \theta)(1 -\sin \theta) } + \frac{ 1(1 +\sin \theta) }{ (1 -\sin \theta)(1 +\sin \theta) }\]


\[\frac{ 1 -\sin \theta }{ (1 +\sin \theta)(1 -\sin \theta) } + \frac{ 1 +\sin \theta }{ (1 -\sin \theta)(1 +\sin \theta) }\]


\[\frac{ 1 -\sin \theta }{ (1 +\sin \theta)(1 -\sin \theta) } + \frac{ 1 +\sin \theta }{ (1 +\sin \theta)(1 -\sin \theta) }\]


\[\frac{ 1 -\sin \theta + 1 +\sin \theta }{ (1 +\sin \theta)(1 -\sin \theta) }\]


\[\frac{ 2 }{ (1 +\sin \theta)(1 -\sin \theta) }\]


\[\frac{ 2 }{ 1^2 -\sin^2 \theta}\]


\[\frac{ 2 }{ \cos^2 \theta}\]


\[2\sec^2 \theta\]


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So


\[\frac{ 1 }{ 1 +\sin \theta } + \frac{ 1 }{ 1 -\sin \theta } = 2\sec^2 \theta\]


is an identity

Answer 27

that is PERFECTLY explained, thank you!!!

Answer 28

you're welcome

Answer 29

I went as far as \[\frac{ 2 }{ (1 - \sin \theta)^{2} }\]

but couldn't solve further :(

Answer 30

now I understand, thank you!

Answer 31

that's great, yw

Answer 32

keep in mind that sin^2 + cos^2 = 1 and this identity is very useful (it also helps you find other identities)

Answer 33

Can you help me really quick with another one? I think it's easier than the previous one.
Here it is:

\[\frac{ \sin ^{2}\theta }{ 1 - \cos \theta }\]

Answer 34

again, sin^2 + cos^2 = 1, so this means..

sin^2 + cos^2 = 1

sin^2 = 1 - cos^2

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For our problem, we can say

\[\frac{ \sin ^{2}\theta }{ 1 - \cos \theta }\]


\[\frac{ 1 - \cos ^{2}\theta }{ 1 - \cos \theta }\]


\[\frac{ 1^2 - \cos ^{2}\theta }{ 1 - \cos \theta }\]


\[\frac{ (1 - \cos\theta)(1 + \cos\theta) }{ 1 - \cos \theta }\]


See what to do from here?

Answer 35

cancel out the (1−cosθ) from the top and bottom, leaving me with (1+cosθ)!!! ? :)

Answer 36

you nailed it

Answer 37

or just 1 + cosθ, but that's the same thing

Answer 38

you're amazing, thank you so so much!! :)

Answer 39

glad to be of help