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Let us know if you need more help with this ...

why did you muliply 1 over cos theta?

it's so interesting how we can do that :p

\[\cos^2(x)=\cos(x) \cdot \cos(x)\]

oh i missed that. thanks guys!!!! :)

do you guys memorize these formulas?

Yes!

I guess I have to do lots of these questions to master this chapter..

tehehehehe aywy thanks so much lifesaverss :)

haha I don't even see the first step...

For
cot x and csc x
I would change them into sines and cosines, can you do this?

......?

for csc x 1 over sin x

No .... for cot theta over csc theta

is it\[{\cot\theta\over\csc\theta-1}={\csc\theta+1\over\cot\theta}\]??

yes

I would change
cot (theta) = cos(theta) / sin (theta)
csc(theta) = 1/sin(theta)
for all cot and csc

yes I did that just now :)

Yes meverett is right, though this one is a little tricky. I had to change both sides.

and multiply each side by 1 over sin theta - 1?

sorry no... let me fix that.

\[{{\cos x\over\sin x}\over{1\over\sin x}-1}={{{1\over\sin x}+1}\over{\cos x\over\sin x}}\]right?

uh... i don't get the 2nd part. you flip them?

no, that's just the right side of the formula with csc written as 1/sin, and cot written as cos/sin.

oh hahahaha gotcha

uhh I am sorry I don't want your head to hurt.. but I don't know how to do it.. :S

no prob, here it is...

we just multiplied the top and bottom of each fraction by sinx

yes!

it should change the denominator in a familiar way...

i got it 'till \[cosx + sinxcosx \over 1-\sin ^{2} x\]

\[\cos ^{2}x\]

so we can divide out our fraction by a cosx to make it...?

yes!!!!!!!!!!!!!!!!!!!!!!!!!!

there ya go :)

THANKS! yayayayayay!

you are really qualified as an awesome tutor! so patient.. :') i wish you the bestest luck !

Turing is awesome!

so are you though, your every each step helped me alot :)

myinininaya :)

Yay myin!

LOVELY! happy NEW year :) bye ttyl

nikvist can u hlp me??