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

1. anonymous

$\frac{1}{1+\sin\theta}=\frac{1-\sin\theta}{1-\sin^2\theta}=\frac{1-\sin\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}-\frac{\sin\theta}{\cos^2\theta}=\sec^2\theta-\frac{\tan\theta}{\cos\theta}$

2. anonymous

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

3. 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

4. anonymous

why did you muliply 1 over cos theta?

5. anonymous

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

6. myininaya

$\cos^2(x)=\cos(x) \cdot \cos(x)$

7. anonymous

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

8. 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

9. anonymous

do you guys memorize these formulas?

10. myininaya

Yes!

11. anonymous

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

12. anonymous

tehehehehe aywy thanks so much lifesaverss :)

13. 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

14. 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 ???

15. anonymous

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

16. anonymous

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

17. anonymous

......?

18. anonymous

for csc x 1 over sin x

19. anonymous

No .... for cot theta over csc theta

20. TuringTest

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

21. anonymous

yes

22. anonymous

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

23. anonymous

yes I did that just now :)

24. TuringTest

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

25. anonymous

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

26. TuringTest

sorry no... let me fix that.

27. TuringTest

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

28. anonymous

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

29. TuringTest

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

30. anonymous

oh hahahaha gotcha

31. 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.

32. anonymous

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

33. TuringTest

no prob, here it is...

34. 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?

35. TuringTest

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

36. anonymous

yes!

37. 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.

38. TuringTest

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

39. anonymous

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

40. 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?

41. anonymous

$\cos ^{2}x$

42. TuringTest

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

43. anonymous

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

44. TuringTest

there ya go :)

45. anonymous

THANKS! yayayayayay!

46. anonymous

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

47. myininaya

Turing is awesome!

48. 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!

49. anonymous

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

50. anonymous

myinininaya :)

51. TuringTest

Yay myin!

52. anonymous

LOVELY! happy NEW year :) bye ttyl

53. AravindG

nikvist can u hlp me??