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

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- 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

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## 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.

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