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Hannah_Ahn
Group Title
Prove the identity.
\[1 \over 1+\sin \theta\]=\[\sec ^{2} \theta\] \[ \tan \theta \over \cos \theta\]
 2 years ago
 2 years ago
Hannah_Ahn Group Title
Prove the identity. \[1 \over 1+\sin \theta\]=\[\sec ^{2} \theta\] \[ \tan \theta \over \cos \theta\]
 2 years ago
 2 years ago

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nikvist Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{1}{1+\sin\theta}=\frac{1\sin\theta}{1\sin^2\theta}=\frac{1\sin\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}\frac{\sin\theta}{\cos^2\theta}=\sec^2\theta\frac{\tan\theta}{\cos\theta}\]
 2 years ago

meverett04 Group TitleBest ResponseYou've already chosen the best response.4
Let us know if you need more help with this ...
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
\[\frac{1}{1+\sin(x)} \cdot \frac{1\sin(x)}{1\sin(x)}=\frac{1\sin(x)}{1\sin^2(x)}=\frac{1\sin(x)}{\cos^2(x)}\] \[=\frac{1}{\cos^2(x)}\frac{\sin(x)}{\cos^2(x)}=\sec^2(x)\frac{\sin(x)}{\cos(x)} \cdot \frac{1}{\cos(x)}\] \[=\sec^2(x)\tan(x) \cdot \frac{1}{\cos(x)}=\sec^2(x)\frac{\tan(x)}{\cos(x)}\] but nikvist already did this i might in a little extra steps if you wanted to see those
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
why did you muliply 1 over cos theta?
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
it's so interesting how we can do that :p
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
\[\cos^2(x)=\cos(x) \cdot \cos(x)\]
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
oh i missed that. thanks guys!!!! :)
 2 years ago

meverett04 Group TitleBest ResponseYou've already chosen the best response.4
I don't think they multiplied 1 over cos theta .... I think they used the Pythagorean Identity sin^2 x + cos^2 x = 1 or 1  sin^2 x = cos^2 x You will use this formula many many many times in trig
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
do you guys memorize these formulas?
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
I guess I have to do lots of these questions to master this chapter..
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
tehehehehe aywy thanks so much lifesaverss :)
 2 years ago

meverett04 Group TitleBest ResponseYou've already chosen the best response.4
I only memorize the ones I see over and over again. For trig, there are just a few that I memorize sin^2 x + cos^2 x = 1 is one of them, but I have used it so many times, it is no longer memorized, I know it like my name meverett lol
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
I am having troubles with every single questions.. wowo.... how do i prove this one.? cot theta over csc theta  1 equals csc theta + 1 over cot theta ???
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
haha I don't even see the first step...
 2 years ago

meverett04 Group TitleBest ResponseYou've already chosen the best response.4
For cot x and csc x I would change them into sines and cosines, can you do this?
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
for csc x 1 over sin x
 2 years ago

meverett04 Group TitleBest ResponseYou've already chosen the best response.4
No .... for cot theta over csc theta
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
is it\[{\cot\theta\over\csc\theta1}={\csc\theta+1\over\cot\theta}\]??
 2 years ago

meverett04 Group TitleBest ResponseYou've already chosen the best response.4
I would change cot (theta) = cos(theta) / sin (theta) csc(theta) = 1/sin(theta) for all cot and csc
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
yes I did that just now :)
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
Yes meverett is right, though this one is a little tricky. I had to change both sides.
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
and multiply each side by 1 over sin theta  1?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
sorry no... let me fix that.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
\[{{\cos x\over\sin x}\over{1\over\sin x}1}={{{1\over\sin x}+1}\over{\cos x\over\sin x}}\]right?
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
uh... i don't get the 2nd part. you flip them?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
no, that's just the right side of the formula with csc written as 1/sin, and cot written as cos/sin.
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
oh hahahaha gotcha
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I changed both sides to their elementary functions is all. I think there are too many sines in this equation; makes my head hurt. But it looks like if we multiply each side by\[\frac{\sin x}{\sin x}\]everything will be a bit simpler, so try that. See if you can write what you get.
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
uhh I am sorry I don't want your head to hurt.. but I don't know how to do it.. :S
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
no prob, here it is...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{\frac{\cos x}{\sin x}}{\frac{1}{\sin x}1}\frac{\sin x}{\sin x}=\frac{\frac{1}{\sin x}+1}{\frac{\cos x}{\sin x}}\frac{\sin x}{\sin x}\]\[\frac{\cos x}{1\sin x}=\frac{1+\sin x}{\cos x}\]reasonable?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
we just multiplied the top and bottom of each fraction by sinx
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
Ok, so now I'm going to say that my mission is to get the left to look like the right. I'm not going to change the right anymore, just try to get the left to look like it. So see if you can multiply the left by (1+sinx)/(1+sinx). That should make you see something.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
it should change the denominator in a familiar way...
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
i got it 'till \[cosx + sinxcosx \over 1\sin ^{2} x\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
That's good. You didn't need to multiply out the top, you could have left it as cosx(1+sinx). You'll see why in this step. Look at the denominator. What is that identity?
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
\[\cos ^{2}x\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
so we can divide out our fraction by a cosx to make it...?
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
yes!!!!!!!!!!!!!!!!!!!!!!!!!!
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
there ya go :)
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
THANKS! yayayayayay!
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
you are really qualified as an awesome tutor! so patient.. :') i wish you the bestest luck !
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Turing is awesome!
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
so what was the strategy? 1)convert all tan, cot, sec, and csc into terms of sin and cos. 2)Simplify if possible. Remember that mertsj was right, you cannot "divide both sides" or "multiply both sides" in identities. You need to work each side separately. 3)Look for ways make things like (1sin^2x) so we can change it to cos^2x. That was all my strategy was, plus practice. Happy New Year!
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
so are you though, your every each step helped me alot :)
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
myinininaya :)
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
Yay myin!
 2 years ago

Hannah_Ahn Group TitleBest ResponseYou've already chosen the best response.1
LOVELY! happy NEW year :) bye ttyl
 2 years ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
nikvist can u hlp me??
 2 years ago
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