## 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

1. nikvist Group Title

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

2. meverett04 Group Title

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

3. myininaya Group Title

$\frac{1}{1+\sin(x)} \cdot \frac{1-\sin(x)}{1-\sin(x)}=\frac{1-\sin(x)}{1-\sin^2(x)}=\frac{1-\sin(x)}{\cos^2(x)}$ $=\frac{1}{\cos^2(x)}-\frac{\sin(x)}{\cos^2(x)}=\sec^2(x)-\frac{\sin(x)}{\cos(x)} \cdot \frac{1}{\cos(x)}$ $=\sec^2(x)-\tan(x) \cdot \frac{1}{\cos(x)}=\sec^2(x)-\frac{\tan(x)}{\cos(x)}$ but nikvist already did this i might in a little extra steps if you wanted to see those

4. Hannah_Ahn Group Title

why did you muliply 1 over cos theta?

5. Hannah_Ahn Group Title

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

6. myininaya Group Title

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

7. Hannah_Ahn Group Title

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

8. meverett04 Group Title

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. Hannah_Ahn Group Title

do you guys memorize these formulas?

10. myininaya Group Title

Yes!

11. Hannah_Ahn Group Title

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

12. Hannah_Ahn Group Title

tehehehehe aywy thanks so much lifesaverss :)

13. meverett04 Group Title

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. Hannah_Ahn Group Title

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. Hannah_Ahn Group Title

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

16. meverett04 Group Title

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

17. Hannah_Ahn Group Title

......?

18. Hannah_Ahn Group Title

for csc x 1 over sin x

19. meverett04 Group Title

No .... for cot theta over csc theta

20. TuringTest Group Title

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

21. Hannah_Ahn Group Title

yes

22. meverett04 Group Title

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

23. Hannah_Ahn Group Title

yes I did that just now :)

24. TuringTest Group Title

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

25. Hannah_Ahn Group Title

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

26. TuringTest Group Title

sorry no... let me fix that.

27. TuringTest Group Title

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

28. Hannah_Ahn Group Title

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

29. TuringTest Group Title

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

30. Hannah_Ahn Group Title

oh hahahaha gotcha

31. TuringTest Group Title

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. Hannah_Ahn Group Title

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

33. TuringTest Group Title

no prob, here it is...

34. TuringTest Group Title

$\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 Group Title

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

36. Hannah_Ahn Group Title

yes!

37. TuringTest Group Title

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 Group Title

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

39. Hannah_Ahn Group Title

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

40. TuringTest Group Title

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. Hannah_Ahn Group Title

$\cos ^{2}x$

42. TuringTest Group Title

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

43. Hannah_Ahn Group Title

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

44. TuringTest Group Title

there ya go :)

45. Hannah_Ahn Group Title

THANKS! yayayayayay!

46. Hannah_Ahn Group Title

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

47. myininaya Group Title

Turing is awesome!

48. TuringTest Group Title

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. Hannah_Ahn Group Title

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

50. Hannah_Ahn Group Title

myinininaya :)

51. TuringTest Group Title

Yay myin!

52. Hannah_Ahn Group Title

LOVELY! happy NEW year :) bye ttyl

53. AravindG Group Title

nikvist can u hlp me??