anonymous
  • anonymous
VERIFY THE IDENTITY: 1/csc(x-1) + 1/csc(x+1)=2/csc(x)-sin(x)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
freckles
  • freckles
question is it really: \[\frac{1}{\csc(x-1)}+\frac{1}{\csc(x+1)}=\frac{2}{\csc(x)}-\sin(x)\]
anonymous
  • anonymous
close, only it says 2/csc(x)-sin(x) not sure if that is the same thing or not?
freckles
  • freckles
isn't that what I typed for the right hand side?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

freckles
  • freckles
did you mean to write 2/(csc(x)-sin(x)) instead?
anonymous
  • anonymous
yes
freckles
  • freckles
\[\frac{1}{\csc(x-1)}+\frac{1}{\csc(x+1)}=\frac{2}{\csc(x)-\sin(x)}\]
anonymous
  • anonymous
yes that is correct
freckles
  • freckles
that doesn't seem like an identity
freckles
  • freckles
try inputting pi/6
freckles
  • freckles
both sides aren't the same
anonymous
  • anonymous
Ok that is what I kept coming up with, I just wanted to verify that it was not an identity, thank you!
freckles
  • freckles
did you mean to write \[\frac{1}{\csc(x)-1}+\frac{1}{\csc(x)+1}=\frac{2}{\csc(x)-\sin(x)}\] this is an identity
anonymous
  • anonymous
I cannot figure out how it is an identity, however....
freckles
  • freckles
so is that not what you are trying to prove is an identity ? it really is the first equation you wrote?
freckles
  • freckles
If you really meant to write that last thing I wrote, try combining the fractions on the left hand side first
anonymous
  • anonymous
I think I am typing it wrong, this is exactly how it is written on the worksheet: \[\frac{ 1 }{ \csc x-1 }+\frac{ 1 }{ \csc x+1 }=\frac{ 2 }{ \csc x-\sin x }\]
freckles
  • freckles
then all you have to do is divide both top and bottom by csc(x)
freckles
  • freckles
2 steps really: combine fractions divide top and bot by csc(x)
anonymous
  • anonymous
How can you combine the fractions when the denominator are not the same? I am not sure how to make them the same
UnkleRhaukus
  • UnkleRhaukus
\[LHS=\frac{1}{\csc(x)-1}+\frac{1}{\csc(x)+1}\\ =\frac{1}{\csc(x)-1}\times\frac{\csc(x)+1}{\csc(x)+1}+\frac{1}{\csc(x)+1}\times\frac{\csc(x)-1}{\csc(x)-1}\\ =\frac{2\csc(x)}{\csc^2(x)-1}\\ =\] now divide both numerator and denominator by csc
anonymous
  • anonymous
Like this? \[\frac{ 2 }{ \csc x-1}\]
UnkleRhaukus
  • UnkleRhaukus
\[\frac{ 2 }{ \csc x-\tfrac1{\csc(x)}}\]
anonymous
  • anonymous
ohh of course. I understand now. Thank you so much!
UnkleRhaukus
  • UnkleRhaukus
all clear? do you understand how we combined those fractions at the start?
anonymous
  • anonymous
yes, I kept trying to do it that way but for some reason I was doing the algebra wrong....I need to practice more

Looking for something else?

Not the answer you are looking for? Search for more explanations.