anonymous
  • anonymous
Verify the identity tan(θ)cot(θ)-sin^2θ=cos^2θ
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
bahrom7893
  • bahrom7893
Hey, 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
  • bahrom7893
Bear in mind: tan(t) = Sin(t)/Cos(t) and Cot(t) = Cos(t)/Sin(t)
bahrom7893
  • bahrom7893
insert those into your equation: [Sin(t)/Cos(t)]*[Cos(t)/Sin(t)] - Sin^2t = Cos^2t Simplify: 1 - Sin^2t = Cos^2t

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bahrom7893
  • bahrom7893
Well actually Tan*Cot = 1, but I showed you where that one came from.
bahrom7893
  • bahrom7893
So 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
anonymous
  • anonymous
Ok 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
  • anonymous
Ok 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
  • anonymous
The 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
  • anonymous
ok I guess i'm not clear on what csc is?
bahrom7893
  • bahrom7893
Csc is Cosecant = 1/Sin
anonymous
  • anonymous
oh ok this is making more sense now! Thank you!
anonymous
  • anonymous
ok sorry one more time what is sin equal to then?
bahrom7893
  • bahrom7893
Sin and Cos are basic units, everything can be broken down into them
bahrom7893
  • bahrom7893
u can't break those down
anonymous
  • anonymous
ok 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
  • bahrom7893
in the first parenthesis is it 1/(1+Sinx), i mean is 1+sinx in the bottom?
anonymous
  • anonymous
yes sorry about that
bahrom7893
  • bahrom7893
and btw is the 1+Cscx in the denominator too? always put parenthesis
anonymous
  • anonymous
Yes sorry about that
bahrom7893
  • bahrom7893
k workin on it now
anonymous
  • anonymous
Thanks
bahrom7893
  • bahrom7893
1/(1+Sinx) + 1/(1+Cscx) = 1/(1+Sinx) + 1/(1 + [1/Sinx])
bahrom7893
  • bahrom7893
1/(1+Sinx) + {1 / [(Sinx+1)/Sinx]} = 1/(1+Sinx) + Sinx/(1+Sinx)
bahrom7893
  • bahrom7893
Now common denominator as you can see is 1+Sinx => You can rewrite as: (1+Sinx)/(1+Sinx) = 1
anonymous
  • anonymous
which narrows down to 1=1 right
bahrom7893
  • bahrom7893
yeah
anonymous
  • anonymous
awesome thanks
anonymous
  • anonymous
ok what if I have ((1-2sin^2(θ))/(sin(θ)cos(θ)))=cot(θ)-tan(θ) I have (1-2sin^2θ)/(sinθ(cosθ/sinθ)=(cotθ-(sinθ/cosθ)? where do I go from there?
bahrom7893
  • bahrom7893
WOW lol
anonymous
  • anonymous
is that wrong?
bahrom7893
  • bahrom7893
no its long lol
anonymous
  • anonymous
oh well the bottom line is what I plugged in....
bahrom7893
  • bahrom7893
hey can u write it out on a piece of paper and take a picture of it?
anonymous
  • anonymous
Uhm ya I do online school so it's online on a word document ....
bahrom7893
  • bahrom7893
okay email that to me
bahrom7893
  • bahrom7893
anonymous
  • anonymous
ok...
anonymous
  • anonymous
ok just sent it
bahrom7893
  • bahrom7893
k lookin at it
anonymous
  • anonymous
?
bahrom7893
  • bahrom7893
workin on it
bahrom7893
  • bahrom7893
[1 - 2 Sin^2(t)] / [Sint*Cost] = Cost/Sint - Sint/Cost
bahrom7893
  • bahrom7893
[1 - 2 Sin^2(t)] / [Sint*Cost] = [Cos^2(t) - Sin^2(t)]/[Sint*Cost]
bahrom7893
  • bahrom7893
subtract [Cos^2(t) - Sin^2(t)]/[Sint*Cost] from both sides
bahrom7893
  • bahrom7893
[1-2Sin^2(t)]/[Sint*Cost] - [Cos^2(t) - Sin^2(t)]/[Sint*Cost] = 0
anonymous
  • anonymous
Ok I'm lost...
bahrom7893
  • bahrom7893
where?
anonymous
  • anonymous
Sorry I'm probably frustrating you.... how did you get it equal to 0?
bahrom7893
  • bahrom7893
I subtracted [Cos^2(t) - Sin^2(t)]/[Sint*Cost] from both sides
bahrom7893
  • bahrom7893
[Cos^2(t) - Sin^2(t)]/[Sint*Cost] - [Cos^2(t) - Sin^2(t)]/[Sint*Cost] = 0 on the left
bahrom7893
  • bahrom7893
get it?
bahrom7893
  • bahrom7893
I meant on the right side: [Cos^2(t) - Sin^2(t)]/[Sint*Cost] - [Cos^2(t) - Sin^2(t)]/[Sint*Cost] is 0
anonymous
  • anonymous
Oh oh ok.. Ihad the right side... I get it thank you I'm messing with the number 10 problem now
bahrom7893
  • bahrom7893
k so
bahrom7893
  • bahrom7893
[1-2Sin^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
anonymous
  • anonymous
ok
bahrom7893
  • bahrom7893
Simplify the top: ( 1 - Sin^2(t) - Cos^2(t)) / (Sint*Cost) = 0
bahrom7893
  • bahrom7893
notice that the top is also: ( 1 - [Sin^2(t) + Cos^2(t)]) / (Sint*Cost) = 0
bahrom7893
  • bahrom7893
(1-1)/(Sint*Cost) = 0 0/(Sint*Cost) = 0 0 = 0 VOILA!
bahrom7893
  • bahrom7893
and can u become my fan lol, im tryin to get more fans
bahrom7893
  • bahrom7893
For the last one, square both sides
anonymous
  • anonymous
Ha ha I already did! Thank you so much if I could be your fan more than once I would thank you so much!
bahrom7893
  • bahrom7893
LOl its fine
bahrom7893
  • bahrom7893
(1+Sint)/(1-Sint) = (1+Sint)^2 / Cos^2(t), u can ignore the abs value cuz Cos^2 is always positive
anonymous
  • anonymous
ahh wait ok just checking the stuff youjust showed me.....that was #4 right?
bahrom7893
  • bahrom7893
yeah
anonymous
  • anonymous
ok I had it down there but wasn't sure... ok I'm following ya now
bahrom7893
  • bahrom7893
okay can i work on ur last one in a couple of mins? i promised to help someone
anonymous
  • anonymous
Ya no problem I'll mess around with it and see how far I get thanks
bahrom7893
  • bahrom7893
okay how far r u?
anonymous
  • anonymous
not much farther... I started where we left off but got way lost... I think I started dividing things that shouldnt be divided
anonymous
  • anonymous
This is what I have.... wait did you say square both sides?
bahrom7893
  • bahrom7893
yeah
bahrom7893
  • bahrom7893
you'll get: (1+Sint)/(1-Sint) = (1+Sint)^2/Cos^2(t)
anonymous
  • anonymous
ok I have that
bahrom7893
  • bahrom7893
Okay now move to th(1+Sint)/(1-Sint) - (1+Sint)^2/Cos^2(t) = 0
bahrom7893
  • bahrom7893
Divide everything by (1+Sint)
anonymous
  • anonymous
ok so then... (1-sint)-cos^2(t)?
bahrom7893
  • bahrom7893
no
bahrom7893
  • bahrom7893
1/(1-Sint) = (1+Sint)/Cos^2(t)
anonymous
  • anonymous
ugh... alright oh ok
anonymous
  • anonymous
sorry I didn't mean to take up this much of your time
bahrom7893
  • bahrom7893
1 - Sin(t) = (Cos^2(t))/(1+Sin(t))
bahrom7893
  • bahrom7893
Multiply both sides by (1+Sin(t)): (1-Sint) (1+Sint) = Cos^2(t)
bahrom7893
  • bahrom7893
Recognize 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
  • anonymous
I'm really sorry... thank you and good luck with your fan contest! thank you so very much!
bahrom7893
  • bahrom7893
lol its not a contest, its just for fun lol and ur welcome!
anonymous
  • anonymous
I wish I understood it more but that's the faultyness of online school... I get the jist! thanks
bahrom7893
  • bahrom7893
no 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
  • anonymous
I hear ya! Night thanks again!

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