Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

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
Bear in mind: tan(t) = Sin(t)/Cos(t) and Cot(t) = Cos(t)/Sin(t)
insert those into your equation: [Sin(t)/Cos(t)]*[Cos(t)/Sin(t)] - Sin^2t = Cos^2t Simplify: 1 - Sin^2t = Cos^2t

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question