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anonymous
 4 years ago
prove the identity
csc theta over tan theta + cot theta = cos theta
anonymous
 4 years ago
prove the identity csc theta over tan theta + cot theta = cos theta

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ \frac{1}{sin {\theta}}}{tan{\theta}}+\frac{1}{tan{\theta}}=cos{\theta}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am sorry, how did that happen?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry I will write it up first and post it with steps. Just a moment

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0does it mean that (1/sin x)^2+(1/sin x)=1?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if you divide the cos theta from both sides

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I still would appreciate to see the steps plz :s

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but why? let x=pi/6 then 4+2=1??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes I'm having that issue too. Are you sure it's written down right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\csc \theta \over \tan \theta +\cot \theta\] = [\cos \theta\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I see now, next time put the tan theta + cot theta in brackets to make it clearer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0multiply the LHS by sinx cosx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0gotcha. sorry about that.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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
 4 years ago
Best ResponseYou've already chosen the best response.0im sorry it should be theta...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why did you multiply each sides by sinx cosx? how could u tell?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1/sin{\theta}}{tan{\theta}+1/tan{\theta}}=cos{\theta}\]\[1/sin{\theta}=tan{\theta}cos{\theta}+cos{\theta}/tan{\theta}\]\[1/sin{\theta}=sin{\theta}+cos^2{\theta}/sin{\theta}\]\[1/sin{\theta}=sin^2{\theta}/sin{\theta}+cos^2{\theta}/sin{\theta}\]\[1/sin{\theta}=1/sin{\theta}\]\[1=1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here's a different way to go about it :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because there are sinx cosx in the denominator

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my 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
 4 years ago
Best ResponseYou've already chosen the best response.0zed; yur way is something to think about, and it's cool. yet i won't b able to use your method

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Zed you shouldnt give out a prove like this, you should write LHS=...=....=...=RHS

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yifian; i don't really get how you got there.. :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i can't figure it out

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is there other way to do it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks 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
 4 years ago
Best ResponseYou've already chosen the best response.0sorry that second last line was supposed to be cosx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah that is a good prove except that you substitute theta for x..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah i got tired of writing theta out

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01 over cos x isn't same as cos x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well thanks for your time :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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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