anonymous
  • anonymous
Prove the identity. \[1 \over 1+\sin \theta\]=\[\sec ^{2} \theta\] -\[ \tan \theta \over \cos \theta\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
nikvist
  • nikvist
\[\frac{1}{1+\sin\theta}=\frac{1-\sin\theta}{1-\sin^2\theta}=\frac{1-\sin\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}-\frac{\sin\theta}{\cos^2\theta}=\sec^2\theta-\frac{\tan\theta}{\cos\theta}\]
anonymous
  • anonymous
Let us know if you need more help with this ...
myininaya
  • myininaya
\[\frac{1}{1+\sin(x)} \cdot \frac{1-\sin(x)}{1-\sin(x)}=\frac{1-\sin(x)}{1-\sin^2(x)}=\frac{1-\sin(x)}{\cos^2(x)}\] \[=\frac{1}{\cos^2(x)}-\frac{\sin(x)}{\cos^2(x)}=\sec^2(x)-\frac{\sin(x)}{\cos(x)} \cdot \frac{1}{\cos(x)}\] \[=\sec^2(x)-\tan(x) \cdot \frac{1}{\cos(x)}=\sec^2(x)-\frac{\tan(x)}{\cos(x)}\] but nikvist already did this i might in a little extra steps if you wanted to see those

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anonymous
  • anonymous
why did you muliply 1 over cos theta?
anonymous
  • anonymous
it's so interesting how we can do that :p
myininaya
  • myininaya
\[\cos^2(x)=\cos(x) \cdot \cos(x)\]
anonymous
  • anonymous
oh i missed that. thanks guys!!!! :)
anonymous
  • anonymous
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
anonymous
  • anonymous
do you guys memorize these formulas?
myininaya
  • myininaya
Yes!
anonymous
  • anonymous
I guess I have to do lots of these questions to master this chapter..
anonymous
  • anonymous
tehehehehe aywy thanks so much lifesaverss :)
anonymous
  • anonymous
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
anonymous
  • anonymous
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 ???
anonymous
  • anonymous
haha I don't even see the first step...
anonymous
  • anonymous
For cot x and csc x I would change them into sines and cosines, can you do this?
anonymous
  • anonymous
......?
anonymous
  • anonymous
for csc x 1 over sin x
anonymous
  • anonymous
No .... for cot theta over csc theta
TuringTest
  • TuringTest
is it\[{\cot\theta\over\csc\theta-1}={\csc\theta+1\over\cot\theta}\]??
anonymous
  • anonymous
yes
anonymous
  • anonymous
I would change cot (theta) = cos(theta) / sin (theta) csc(theta) = 1/sin(theta) for all cot and csc
anonymous
  • anonymous
yes I did that just now :)
TuringTest
  • TuringTest
Yes meverett is right, though this one is a little tricky. I had to change both sides.
anonymous
  • anonymous
and multiply each side by 1 over sin theta - 1?
TuringTest
  • TuringTest
sorry no... let me fix that.
TuringTest
  • TuringTest
\[{{\cos x\over\sin x}\over{1\over\sin x}-1}={{{1\over\sin x}+1}\over{\cos x\over\sin x}}\]right?
anonymous
  • anonymous
uh... i don't get the 2nd part. you flip them?
TuringTest
  • TuringTest
no, that's just the right side of the formula with csc written as 1/sin, and cot written as cos/sin.
anonymous
  • anonymous
oh hahahaha gotcha
TuringTest
  • TuringTest
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.
anonymous
  • anonymous
uhh I am sorry I don't want your head to hurt.. but I don't know how to do it.. :S
TuringTest
  • TuringTest
no prob, here it is...
TuringTest
  • TuringTest
\[\frac{\frac{\cos x}{\sin x}}{\frac{1}{\sin x}-1}\frac{\sin x}{\sin x}=\frac{\frac{1}{\sin x}+1}{\frac{\cos x}{\sin x}}\frac{\sin x}{\sin x}\]\[\frac{\cos x}{1-\sin x}=\frac{1+\sin x}{\cos x}\]reasonable?
TuringTest
  • TuringTest
we just multiplied the top and bottom of each fraction by sinx
anonymous
  • anonymous
yes!
TuringTest
  • TuringTest
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.
TuringTest
  • TuringTest
it should change the denominator in a familiar way...
anonymous
  • anonymous
i got it 'till \[cosx + sinxcosx \over 1-\sin ^{2} x\]
TuringTest
  • TuringTest
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?
anonymous
  • anonymous
\[\cos ^{2}x\]
TuringTest
  • TuringTest
so we can divide out our fraction by a cosx to make it...?
anonymous
  • anonymous
yes!!!!!!!!!!!!!!!!!!!!!!!!!!
TuringTest
  • TuringTest
there ya go :)
anonymous
  • anonymous
THANKS! yayayayayay!
anonymous
  • anonymous
you are really qualified as an awesome tutor! so patient.. :') i wish you the bestest luck !
myininaya
  • myininaya
Turing is awesome!
TuringTest
  • TuringTest
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 (1-sin^2x) so we can change it to cos^2x. That was all my strategy was, plus practice. Happy New Year!
anonymous
  • anonymous
so are you though, your every each step helped me alot :)
anonymous
  • anonymous
myinininaya :)
TuringTest
  • TuringTest
Yay myin!
anonymous
  • anonymous
LOVELY! happy NEW year :) bye ttyl
AravindG
  • AravindG
nikvist can u hlp me??

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