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

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 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

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?

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

k so

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

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?

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)

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

I hear ya! Night thanks again!