Here's the question you clicked on:
Hannah_Ahn
Prove the identity. \[1 \over 1+\sin \theta\]=\[\sec ^{2} \theta\] -\[ \tan \theta \over \cos \theta\]
\[\frac{1}{1+\sin\theta}=\frac{1-\sin\theta}{1-\sin^2\theta}=\frac{1-\sin\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}-\frac{\sin\theta}{\cos^2\theta}=\sec^2\theta-\frac{\tan\theta}{\cos\theta}\]
Let us know if you need more help with this ...
\[\frac{1}{1+\sin(x)} \cdot \frac{1-\sin(x)}{1-\sin(x)}=\frac{1-\sin(x)}{1-\sin^2(x)}=\frac{1-\sin(x)}{\cos^2(x)}\] \[=\frac{1}{\cos^2(x)}-\frac{\sin(x)}{\cos^2(x)}=\sec^2(x)-\frac{\sin(x)}{\cos(x)} \cdot \frac{1}{\cos(x)}\] \[=\sec^2(x)-\tan(x) \cdot \frac{1}{\cos(x)}=\sec^2(x)-\frac{\tan(x)}{\cos(x)}\] but nikvist already did this i might in a little extra steps if you wanted to see those
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!!!! :)
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
do you guys memorize these formulas?
I guess I have to do lots of these questions to master this chapter..
tehehehehe aywy thanks so much lifesaverss :)
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
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 ???
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\[{\cot\theta\over\csc\theta-1}={\csc\theta+1\over\cot\theta}\]??
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.
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.
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...
\[\frac{\frac{\cos x}{\sin x}}{\frac{1}{\sin x}-1}\frac{\sin x}{\sin x}=\frac{\frac{1}{\sin x}+1}{\frac{\cos x}{\sin x}}\frac{\sin x}{\sin x}\]\[\frac{\cos x}{1-\sin x}=\frac{1+\sin x}{\cos x}\]reasonable?
we just multiplied the top and bottom of each fraction by sinx
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.
it should change the denominator in a familiar way...
i got it 'till \[cosx + sinxcosx \over 1-\sin ^{2} x\]
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?
so we can divide out our fraction by a cosx to make it...?
yes!!!!!!!!!!!!!!!!!!!!!!!!!!
THANKS! yayayayayay!
you are really qualified as an awesome tutor! so patient.. :') i wish you the bestest luck !
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!
so are you though, your every each step helped me alot :)
LOVELY! happy NEW year :) bye ttyl
nikvist can u hlp me??