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anonymous
 5 years ago
Verify the identity
tan(θ)cot(θ)sin^2θ=cos^2θ
anonymous
 5 years ago
Verify the identity tan(θ)cot(θ)sin^2θ=cos^2θ

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bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Hey, I don't feel like copying and pasting, so I'll just use t instead of theta. tan(t)cot(t)Sin^2t = Cos^2t. I rewrote it in terms of t, so that it;s easier for me to type

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Bear in mind: tan(t) = Sin(t)/Cos(t) and Cot(t) = Cos(t)/Sin(t)

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0insert those into your equation: [Sin(t)/Cos(t)]*[Cos(t)/Sin(t)]  Sin^2t = Cos^2t Simplify: 1  Sin^2t = Cos^2t

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Well actually Tan*Cot = 1, but I showed you where that one came from.

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0So now remember the identity: Sin^2 + Cos^2 = 1, well that's the case; move Sin^2t to the right: 1 = Cos^2t + Sin^2t 1 = 1 <identity is verified

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok so I basically am just plugging in what they equal for example tan(t)=sin(t)/cos(t). what if I have a problem with csc... how does that work or what is that equivalent to?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok so I basically am just plugging in what they equal for example tan(t)=sin(t)/cos(t). what if I have a problem with csc... how does that work or what is that equivalent to?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The best thing to do when proving identities is generally to break everything down in terms of sine and cosine. If you know the identities and break everything down, it's just algebra with trig functions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok I guess i'm not clear on what csc is?

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Csc is Cosecant = 1/Sin

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok this is making more sense now! Thank you!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok sorry one more time what is sin equal to then?

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Sin and Cos are basic units, everything can be broken down into them

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0u can't break those down

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so say (1/1+sinx)+(1/1+cscx)=1 (1/1+sinx)+(1/1+(1/sin))=1 ok where do I go from there

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0in the first parenthesis is it 1/(1+Sinx), i mean is 1+sinx in the bottom?

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0and btw is the 1+Cscx in the denominator too? always put parenthesis

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.01/(1+Sinx) + 1/(1+Cscx) = 1/(1+Sinx) + 1/(1 + [1/Sinx])

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.01/(1+Sinx) + {1 / [(Sinx+1)/Sinx]} = 1/(1+Sinx) + Sinx/(1+Sinx)

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Now common denominator as you can see is 1+Sinx => You can rewrite as: (1+Sinx)/(1+Sinx) = 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which narrows down to 1=1 right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok what if I have ((12sin^2(θ))/(sin(θ)cos(θ)))=cot(θ)tan(θ) I have (12sin^2θ)/(sinθ(cosθ/sinθ)=(cotθ(sinθ/cosθ)? where do I go from there?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh well the bottom line is what I plugged in....

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0hey can u write it out on a piece of paper and take a picture of it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Uhm ya I do online school so it's online on a word document ....

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0okay email that to me

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0bahrom.cfo@gmail.com

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0[1  2 Sin^2(t)] / [Sint*Cost] = Cost/Sint  Sint/Cost

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0[1  2 Sin^2(t)] / [Sint*Cost] = [Cos^2(t)  Sin^2(t)]/[Sint*Cost]

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0subtract [Cos^2(t)  Sin^2(t)]/[Sint*Cost] from both sides

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0[12Sin^2(t)]/[Sint*Cost]  [Cos^2(t)  Sin^2(t)]/[Sint*Cost] = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry I'm probably frustrating you.... how did you get it equal to 0?

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0I subtracted [Cos^2(t)  Sin^2(t)]/[Sint*Cost] from both sides

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0[Cos^2(t)  Sin^2(t)]/[Sint*Cost]  [Cos^2(t)  Sin^2(t)]/[Sint*Cost] = 0 on the left

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0I meant on the right side: [Cos^2(t)  Sin^2(t)]/[Sint*Cost]  [Cos^2(t)  Sin^2(t)]/[Sint*Cost] is 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh oh ok.. Ihad the right side... I get it thank you I'm messing with the number 10 problem now

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0[12Sin^2(t)]/[Sint*Cost]  [Cos^2(t)  Sin^2(t)]/[Sint*Cost] = 0 The common denominator is Sint Cost: [1  2*Sin^2(t)  Cos^2(t) + Sin^2(t)] / [Sin(t)*Cos(t)] = 0

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Simplify the top: ( 1  Sin^2(t)  Cos^2(t)) / (Sint*Cost) = 0

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0notice that the top is also: ( 1  [Sin^2(t) + Cos^2(t)]) / (Sint*Cost) = 0

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0(11)/(Sint*Cost) = 0 0/(Sint*Cost) = 0 0 = 0 VOILA!

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0and can u become my fan lol, im tryin to get more fans

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0For the last one, square both sides

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ha ha I already did! Thank you so much if I could be your fan more than once I would thank you so much!

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0(1+Sint)/(1Sint) = (1+Sint)^2 / Cos^2(t), u can ignore the abs value cuz Cos^2 is always positive

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahh wait ok just checking the stuff youjust showed me.....that was #4 right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok I had it down there but wasn't sure... ok I'm following ya now

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0okay can i work on ur last one in a couple of mins? i promised to help someone

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ya no problem I'll mess around with it and see how far I get thanks

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not much farther... I started where we left off but got way lost... I think I started dividing things that shouldnt be divided

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is what I have.... wait did you say square both sides?

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0you'll get: (1+Sint)/(1Sint) = (1+Sint)^2/Cos^2(t)

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Okay now move to th(1+Sint)/(1Sint)  (1+Sint)^2/Cos^2(t) = 0

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Divide everything by (1+Sint)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so then... (1sint)cos^2(t)?

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.01/(1Sint) = (1+Sint)/Cos^2(t)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry I didn't mean to take up this much of your time

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.01  Sin(t) = (Cos^2(t))/(1+Sin(t))

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Multiply both sides by (1+Sin(t)): (1Sint) (1+Sint) = Cos^2(t)

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0Recognize the left side as: 1  Sin^2(t) = Cos^2(t) 1 = Sin^2(t) + Cos^2(t) 1 = 1 <FINALLY DONE WITH UR HW!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm really sorry... thank you and good luck with your fan contest! thank you so very much!

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0lol its not a contest, its just for fun lol and ur welcome!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I wish I understood it more but that's the faultyness of online school... I get the jist! thanks

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0no im off to bed nite. i woke up at 6 am today, left at 6:30 for college got home around 5:30pm, im really tired

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I hear ya! Night thanks again!
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