prove the identity
csc theta over tan theta + cot theta = cos theta

- anonymous

prove the identity
csc theta over tan theta + cot theta = cos theta

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

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

- anonymous

\[\frac{ \frac{1}{sin {\theta}}}{tan{\theta}}+\frac{1}{tan{\theta}}=cos{\theta}\]

- anonymous

I am sorry, how did that happen?

- anonymous

Sorry I will write it up first and post it with steps. Just a moment

Looking for something else?

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

## More answers

- anonymous

Sweet, thanks!

- anonymous

does it mean that (1/sin x)^2+(1/sin x)=1?

- anonymous

if you divide the cos theta from both sides

- anonymous

Yes correct

- anonymous

I still would appreciate to see the steps plz :s

- anonymous

but why? let x=pi/6 then 4+2=1??

- anonymous

yes I'm having that issue too. Are you sure it's written down right?

- anonymous

\[\csc \theta \over \tan \theta +\cot \theta\]
=
[\cos \theta\]

- anonymous

I see now, next time put the tan theta + cot theta in brackets to make it clearer.

- anonymous

multiply the LHS by sinx cosx

- anonymous

gotcha. sorry about that.

- anonymous

i mean csc x /(tan x + cot x)= cscx sinx cosx / (tanx sinx cosx+cotx sinx cosx)=cosx/(sin^2 x+cos^2 x)=cosx

- anonymous

im sorry it should be theta...

- anonymous

why did you multiply each sides by sinx cosx? how could u tell?

- anonymous

\[\frac{1/sin{\theta}}{tan{\theta}+1/tan{\theta}}=cos{\theta}\]\[1/sin{\theta}=tan{\theta}cos{\theta}+cos{\theta}/tan{\theta}\]\[1/sin{\theta}=sin{\theta}+cos^2{\theta}/sin{\theta}\]\[1/sin{\theta}=sin^2{\theta}/sin{\theta}+cos^2{\theta}/sin{\theta}\]\[1/sin{\theta}=1/sin{\theta}\]\[1=1\]

- anonymous

Here's a different way to go about it :)

- anonymous

because there are sinx cosx in the denominator

- anonymous

my teacher doesn't let me combite the left side and the right side.. in this case i would need to make the left side cos theta. :S

- anonymous

zed; yur way is something to think about, and it's cool.
yet i won't b able to use your method

- anonymous

@Zed you shouldnt give out a prove like this, you should write LHS=...=....=...=RHS

- anonymous

yifian; i don't really get how you got there.. :(

- anonymous

i can't figure it out

- anonymous

which step?

- anonymous

is there other way to do it?

- anonymous

Thanks Yifan, I haven't done proof in a long time.
\[LHS = \frac{1/sinx}{sinx/cosx+cosx/sinx}\]
\[=\frac{1/sinx}{(sin^2x+cos^2x)/sinxcosx}\]\[=\frac{1/sinx}{1/sinxcosx}\]\[=\frac{1}{cosx}\]\[=RHS\]

- anonymous

sorry that second last line was supposed to be cosx

- anonymous

yeah that is a good prove except that you substitute theta for x..

- anonymous

yeah i got tired of writing theta out

- anonymous

thanks Yifan

- anonymous

1 over cos x isn't same as cos x

- anonymous

It was a typo

- anonymous

i got that too.

- anonymous

well thanks for your time :)

- anonymous

i still don't get it . for some reason, im very confused. i will ask my teacher and I will let you know if there is a simpler way to do it if you are interested :P g'nite.

Looking for something else?

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