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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{1+\sin\theta}=\frac{1\sin\theta}{1\sin^2\theta}=\frac{1\sin\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}\frac{\sin\theta}{\cos^2\theta}=\sec^2\theta\frac{\tan\theta}{\cos\theta}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let us know if you need more help with this ...

myininaya
 4 years ago
Best 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why did you muliply 1 over cos theta?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it's so interesting how we can do that :p

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh i missed that. thanks guys!!!! :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do you guys memorize these formulas?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I guess I have to do lots of these questions to master this chapter..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0tehehehehe aywy thanks so much lifesaverss :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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 ???

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haha I don't even see the first step...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For cot x and csc x I would change them into sines and cosines, can you do this?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for csc x 1 over sin x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No .... for cot theta over csc theta

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I would change cot (theta) = cos(theta) / sin (theta) csc(theta) = 1/sin(theta) for all cot and csc

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes I did that just now :)

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and multiply each side by 1 over sin theta  1?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0sorry no... let me fix that.

TuringTest
 4 years ago
Best 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0uh... i don't get the 2nd part. you flip them?

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0uhh I am sorry I don't want your head to hurt.. but I don't know how to do it.. :S

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0no prob, here it is...

TuringTest
 4 years ago
Best 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?

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, 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.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i got it 'till \[cosx + sinxcosx \over 1\sin ^{2} x\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0That'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?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes!!!!!!!!!!!!!!!!!!!!!!!!!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you are really qualified as an awesome tutor! so patient.. :') i wish you the bestest luck !

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0so 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!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so are you though, your every each step helped me alot :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0LOVELY! happy NEW year :) bye ttyl

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0nikvist can u hlp me??
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