- anonymous

VERIFY THE IDENTITY:
1/csc(x-1) + 1/csc(x+1)=2/csc(x)-sin(x)

- schrodinger

See more answers at brainly.com

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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- freckles

question is it really:
\[\frac{1}{\csc(x-1)}+\frac{1}{\csc(x+1)}=\frac{2}{\csc(x)}-\sin(x)\]

- anonymous

close, only it says 2/csc(x)-sin(x)
not sure if that is the same thing or not?

- 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

did you mean to write 2/(csc(x)-sin(x)) instead?

- anonymous

yes

- freckles

\[\frac{1}{\csc(x-1)}+\frac{1}{\csc(x+1)}=\frac{2}{\csc(x)-\sin(x)}\]

- anonymous

yes that is correct

- freckles

that doesn't seem like an identity

- freckles

try inputting pi/6

- freckles

both sides aren't the same

- anonymous

Ok that is what I kept coming up with, I just wanted to verify that it was not an identity, thank you!

- 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

I cannot figure out how it is an identity, however....

- freckles

so is that not what you are trying to prove is an identity ? it really is the first equation you wrote?

- freckles

If you really meant to write that last thing I wrote, try combining the fractions on the left hand side first

- 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

then all you have to do is divide both top and bottom by csc(x)

- freckles

2 steps really:
combine fractions
divide top and bot by csc(x)

- 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

\[LHS=\frac{1}{\csc(x)-1}+\frac{1}{\csc(x)+1}\\
=\frac{1}{\csc(x)-1}\times\frac{\csc(x)+1}{\csc(x)+1}+\frac{1}{\csc(x)+1}\times\frac{\csc(x)-1}{\csc(x)-1}\\
=\frac{2\csc(x)}{\csc^2(x)-1}\\
=\]
now divide both numerator and denominator by csc

- anonymous

Like this? \[\frac{ 2 }{ \csc x-1}\]

- UnkleRhaukus

\[\frac{ 2 }{ \csc x-\tfrac1{\csc(x)}}\]

- anonymous

ohh of course. I understand now. Thank you so much!

- UnkleRhaukus

all clear?
do you understand how we combined those fractions at the start?

- 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.