Question

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

Answer 1

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

Answer 2

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

Answer 3

insert those into your equation:
[Sin(t)/Cos(t)]*[Cos(t)/Sin(t)] - Sin^2t = Cos^2t
Simplify:
1 - Sin^2t = Cos^2t

Answer 4

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

Answer 5

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 <identity is verified

Answer 6

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?

Answer 7

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?

Answer 8

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.

Answer 9

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

Answer 10

Csc is Cosecant = 1/Sin

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

u can't break those down

Answer 15

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

Answer 16

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

Answer 17

yes sorry about that

Answer 18

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

Answer 19

Yes sorry about that

Answer 20

k workin on it now

Answer 21

Thanks

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

which narrows down to 1=1 right

Answer 26

yeah

Answer 27

awesome thanks

Answer 28

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?

Answer 29

WOW lol

Answer 30

is that wrong?

Answer 31

no its long lol

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

okay email that to me

Answer 36

bahrom.cfo@gmail.com

Answer 37

ok...

Answer 38

ok just sent it

Answer 39

k lookin at it

Answer 40

?

Answer 41

workin on it

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

Ok I'm lost...

Answer 47

where?

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

get it?

Answer 52

I meant on the right side:
[Cos^2(t) - Sin^2(t)]/[Sint*Cost] - [Cos^2(t) - Sin^2(t)]/[Sint*Cost] is 0

Answer 53

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

Answer 54

k so

Answer 55

[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

Answer 56

ok

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

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

Answer 61

For the last one, square both sides

Answer 62

Ha ha I already did! Thank you so much if I could be your fan more than once I would thank you so much!

Answer 63

LOl its fine

Answer 64

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

Answer 65

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

Answer 66

yeah

Answer 67

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

Answer 68

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

Answer 69

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

Answer 70

okay how far r u?

Answer 71

not much farther... I started where we left off but got way lost... I think I started dividing things that shouldnt be divided

Answer 72

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

Answer 73

yeah

Answer 74

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

Answer 75

ok I have that

Answer 76

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

Answer 77

Divide everything by (1+Sint)

Answer 78

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

Answer 79

no

Answer 80

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

Answer 81

ugh... alright oh ok

Answer 82

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

Answer 83

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

Answer 84

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

Answer 85

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!!!

Answer 86

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

Answer 87

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

Answer 88

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

Answer 89

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

Answer 90

I hear ya! Night thanks again!