Verify the identity
tan(θ)cot(θ)-sin^2θ=cos^2θ

- anonymous

Verify the identity
tan(θ)cot(θ)-sin^2θ=cos^2θ

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- 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

Bear in mind: tan(t) = Sin(t)/Cos(t) and Cot(t) = Cos(t)/Sin(t)

- 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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## More answers

- bahrom7893

Well actually Tan*Cot = 1, but I showed you where that one came from.

- 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

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

ok I guess i'm not clear on what csc is?

- bahrom7893

Csc is Cosecant = 1/Sin

- anonymous

oh ok this is making more sense now! Thank you!

- anonymous

ok sorry one more time what is sin equal to then?

- bahrom7893

Sin and Cos are basic units, everything can be broken down into them

- bahrom7893

u can't break those down

- 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

in the first parenthesis is it 1/(1+Sinx), i mean is 1+sinx in the bottom?

- anonymous

yes sorry about that

- bahrom7893

and btw is the 1+Cscx in the denominator too? always put parenthesis

- anonymous

Yes sorry about that

- bahrom7893

k workin on it now

- anonymous

Thanks

- bahrom7893

1/(1+Sinx) + 1/(1+Cscx) = 1/(1+Sinx) + 1/(1 + [1/Sinx])

- bahrom7893

1/(1+Sinx) + {1 / [(Sinx+1)/Sinx]} = 1/(1+Sinx) + Sinx/(1+Sinx)

- bahrom7893

Now common denominator as you can see is 1+Sinx =>
You can rewrite as:
(1+Sinx)/(1+Sinx) = 1

- anonymous

which narrows down to 1=1 right

- bahrom7893

yeah

- anonymous

awesome thanks

- 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

WOW lol

- anonymous

is that wrong?

- bahrom7893

no its long lol

- anonymous

oh well the bottom line is what I plugged in....

- bahrom7893

hey can u write it out on a piece of paper and take a picture of it?

- anonymous

Uhm ya I do online school so it's online on a word document ....

- bahrom7893

okay email that to me

- bahrom7893

- anonymous

ok...

- anonymous

ok just sent it

- bahrom7893

k lookin at it

- anonymous

?

- bahrom7893

workin on it

- bahrom7893

[1 - 2 Sin^2(t)] / [Sint*Cost] = Cost/Sint - Sint/Cost

- bahrom7893

[1 - 2 Sin^2(t)] / [Sint*Cost] = [Cos^2(t) - Sin^2(t)]/[Sint*Cost]

- bahrom7893

subtract [Cos^2(t) - Sin^2(t)]/[Sint*Cost] from both sides

- bahrom7893

[1-2Sin^2(t)]/[Sint*Cost] - [Cos^2(t) - Sin^2(t)]/[Sint*Cost] = 0

- anonymous

Ok I'm lost...

- bahrom7893

where?

- anonymous

Sorry I'm probably frustrating you.... how did you get it equal to 0?

- bahrom7893

I subtracted [Cos^2(t) - Sin^2(t)]/[Sint*Cost] from both sides

- bahrom7893

[Cos^2(t) - Sin^2(t)]/[Sint*Cost] - [Cos^2(t) - Sin^2(t)]/[Sint*Cost] = 0 on the left

- bahrom7893

get it?

- 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

Oh oh ok.. Ihad the right side... I get it thank you I'm messing with the number 10 problem now

- bahrom7893

k so

- 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

ok

- bahrom7893

Simplify the top:
( 1 - Sin^2(t) - Cos^2(t)) / (Sint*Cost) = 0

- bahrom7893

notice that the top is also:
( 1 - [Sin^2(t) + Cos^2(t)]) / (Sint*Cost) = 0

- bahrom7893

(1-1)/(Sint*Cost) = 0
0/(Sint*Cost) = 0
0 = 0 VOILA!

- bahrom7893

and can u become my fan lol, im tryin to get more fans

- bahrom7893

For the last one, square both sides

- 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

LOl its fine

- bahrom7893

(1+Sint)/(1-Sint) = (1+Sint)^2 / Cos^2(t), u can ignore the abs value cuz Cos^2 is always positive

- anonymous

ahh wait ok just checking the stuff youjust showed me.....that was #4 right?

- bahrom7893

yeah

- anonymous

ok I had it down there but wasn't sure... ok I'm following ya now

- bahrom7893

okay can i work on ur last one in a couple of mins? i promised to help someone

- anonymous

Ya no problem I'll mess around with it and see how far I get thanks

- bahrom7893

okay how far r u?

- 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

This is what I have.... wait did you say square both sides?

- bahrom7893

yeah

- bahrom7893

you'll get:
(1+Sint)/(1-Sint) = (1+Sint)^2/Cos^2(t)

- anonymous

ok I have that

- bahrom7893

Okay now move to th(1+Sint)/(1-Sint) - (1+Sint)^2/Cos^2(t) = 0

- bahrom7893

Divide everything by (1+Sint)

- anonymous

ok so then...
(1-sint)-cos^2(t)?

- bahrom7893

no

- bahrom7893

1/(1-Sint) = (1+Sint)/Cos^2(t)

- anonymous

ugh... alright oh ok

- anonymous

sorry I didn't mean to take up this much of your time

- bahrom7893

1 - Sin(t) = (Cos^2(t))/(1+Sin(t))

- bahrom7893

Multiply both sides by (1+Sin(t)):
(1-Sint) (1+Sint) = Cos^2(t)

- 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

I'm really sorry... thank you and good luck with your fan contest! thank you so very much!

- bahrom7893

lol its not a contest, its just for fun lol and ur welcome!

- anonymous

I wish I understood it more but that's the faultyness of online school... I get the jist! thanks

- 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

I hear ya! Night thanks again!

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