anonymous
  • anonymous
prove the identity csc theta over tan theta + cot theta = cos theta
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\frac{ \frac{1}{sin {\theta}}}{tan{\theta}}+\frac{1}{tan{\theta}}=cos{\theta}\]
anonymous
  • anonymous
I am sorry, how did that happen?
anonymous
  • anonymous
Sorry I will write it up first and post it with steps. Just a moment

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anonymous
  • anonymous
Sweet, thanks!
anonymous
  • anonymous
does it mean that (1/sin x)^2+(1/sin x)=1?
anonymous
  • anonymous
if you divide the cos theta from both sides
anonymous
  • anonymous
Yes correct
anonymous
  • anonymous
I still would appreciate to see the steps plz :s
anonymous
  • anonymous
but why? let x=pi/6 then 4+2=1??
anonymous
  • anonymous
yes I'm having that issue too. Are you sure it's written down right?
anonymous
  • anonymous
\[\csc \theta \over \tan \theta +\cot \theta\] = [\cos \theta\]
anonymous
  • anonymous
I see now, next time put the tan theta + cot theta in brackets to make it clearer.
anonymous
  • anonymous
multiply the LHS by sinx cosx
anonymous
  • anonymous
gotcha. sorry about that.
anonymous
  • 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
  • anonymous
im sorry it should be theta...
anonymous
  • anonymous
why did you multiply each sides by sinx cosx? how could u tell?
anonymous
  • 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
  • anonymous
Here's a different way to go about it :)
anonymous
  • anonymous
because there are sinx cosx in the denominator
anonymous
  • 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
  • anonymous
zed; yur way is something to think about, and it's cool. yet i won't b able to use your method
anonymous
  • anonymous
@Zed you shouldnt give out a prove like this, you should write LHS=...=....=...=RHS
anonymous
  • anonymous
yifian; i don't really get how you got there.. :(
anonymous
  • anonymous
i can't figure it out
anonymous
  • anonymous
which step?
anonymous
  • anonymous
is there other way to do it?
anonymous
  • 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
  • anonymous
sorry that second last line was supposed to be cosx
anonymous
  • anonymous
yeah that is a good prove except that you substitute theta for x..
anonymous
  • anonymous
yeah i got tired of writing theta out
anonymous
  • anonymous
thanks Yifan
anonymous
  • anonymous
1 over cos x isn't same as cos x
anonymous
  • anonymous
It was a typo
anonymous
  • anonymous
i got that too.
anonymous
  • anonymous
well thanks for your time :)
anonymous
  • 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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