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Mathematics 8 Online
OpenStudy (anonymous):

ugghhhhhh

OpenStudy (anonymous):

Not very helpful... What is your problem.

OpenStudy (anonymous):

its um 2/1-cot 22 degrees + cot 23 degrees

OpenStudy (anonymous):

simplify that expression

OpenStudy (jdoe0001):

\(\bf \cfrac{2}{1-cot(22^o)}+cot(23^o)?\)

OpenStudy (anonymous):

yes

OpenStudy (jdoe0001):

somehow I don't see it simplifying further

OpenStudy (anonymous):

You can solve it...

OpenStudy (anonymous):

are you familiar with trigonometric identites?

OpenStudy (anonymous):

uggh i need a math professor or someting

OpenStudy (anonymous):

Well the answer is one but I have no idea how to get there...

OpenStudy (jdoe0001):

@BestRivenNA can you take a quick screenshot of the material and post it here?

OpenStudy (anonymous):

umm well it just says simplify \[\frac{ 2 }{ 1-\cot 22 degrees } +\cot 23 degrees\]

OpenStudy (anonymous):

lol i didnt know how to make a degree sign

OpenStudy (jdoe0001):

right... well... \(\bf \cfrac{2}{1-cot(22^o)}+cot(23^o)\) I don't see it simplifyable further, maybe it's just me.... thus I'd think there's more context to it

OpenStudy (anonymous):

ok :(

OpenStudy (jdoe0001):

so a quick screenshot may show what's expected I'd think

OpenStudy (anonymous):

well in the equation i have there are no paratheses

OpenStudy (jdoe0001):

ok.... so... do you know how to post a picture here? just using the [Attach File] button

OpenStudy (anonymous):

so do i take a picture with my pc cam?

OpenStudy (anonymous):

Yes please.

OpenStudy (anonymous):

well this is what i get it matlab

OpenStudy (anonymous):

um how did you get 1 after the output?

OpenStudy (anonymous):

Unfortunately I dont know...

OpenStudy (anonymous):

omg the answer is 1!

OpenStudy (anonymous):

I never was that great at trig

OpenStudy (anonymous):

i just realized that i could use the calculator and i got 1.00000000006

OpenStudy (jdoe0001):

well... yes.... if that's what's expected... sure

OpenStudy (jdoe0001):

as opposed to a trig identity

OpenStudy (anonymous):

So what are you saying @jdoe0001

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