## anonymous 4 years ago Prove that (1 + secx)(1 - cosx) = sinxtanx??

1. amistre64

write everything in terms of sin and cos to see things better

2. anonymous

so 1 - cosx + secx - cosxsecx = secx - cosx. Now what??

3. anonymous

sec=cos/sin

4. anonymous

no, sec is 1/cos

5. anonymous

hhaha

6. amistre64

i was thinking more of: $1-cos+sec-cos.sec=sin.tan$ $1-cos+sec-cos/cos=sin.sin/cos$ $1-cos+sec-1=sin^2/cos$ $-cos+1/cos=sin^2/cos$ for starters

7. anonymous

but we don't have to do that. we have to work from one side, and expand that to get the other side. It has to be proven.

8. amistre64

this is working one side; I havent changed the left at all except to see it better

9. amistre64

and by left i mean right lol

10. amistre64

$-cos+1/cos=sin^2/cos$ $(-cos^2+1)/cos=sin^2/cos$ $(1-cos^2)/cos=sin^2/cos$ $sin^2/cos=sin^2/cos$

11. anonymous

oh silly me it is a product!! i thought it was a fraction...

12. anonymous

but we are not supposed to even supposed to do ANYTHING with the other side. It's like we're not supposed to know the other side.

13. anonymous

^that's right :P

14. amistre64

15. amistre64

your not spose to know tha tan = sin/cos? your not spose to know that sin.sin = sin^2? your not spose to know what; any trig relations and identities? but somehow transform them into each other?

16. anonymous

not that, yaar. Imagine they give you a question, saying expand ....., leaving it terms of sin/cos/tan. SO we just do that. we expand. but how??

17. anonymous

$(1+\frac{1}{a})(1-a)=\frac{1}{a}-a=\frac{1-a^2}{a}$ now replace a by cosine and you should get it

18. amistre64

since the right side isnt changed; just revert it back into its origanl form and and leave it there; but the process is still the same ...

19. anonymous

it is much easier to do the algebra without the sines and cosines. then put them in at the end.

20. anonymous

but how does that give us sinxtanx??

21. anonymous

but what amistre said is important. it in not a matter of leaving one side untouched. if you can change one side, you can change it back. if you have to hand it in in a certain form, fine, but if you can alter one side you can presumably go backwards

22. amistre64

what? how does sin.sin/cos = sin tan?

23. amistre64

by definition of what sin and tan ARE lol

24. anonymous

ok we did the algebra part, now comes the trig part. replace a by cosine and get $\frac{1-\cos^2(x)}{\cos(x)}$ now come the only trig step, writing $1-\cos^2(x)=\sin^2(x)$ that is the one step that is trig

25. amistre64

can know that 1-cos^2 = sin^2 sat; its forbidden knowledge ;)

26. anonymous

I GOT IT. I love you guys :D

27. anonymous

so you get $\frac{1-\cos^2(x)}{\cos(x)}=\frac{\sin^2(x)}{\cos(x)}=\frac{\sin(x)}{\cos(x)}\times \sin(x)$

28. amistre64

:) good luck

29. anonymous

and we love you!

30. anonymous

okayy, another question. given that tanx = p, find an expression, in terms of p, for cosec^2 x. GO.

31. amistre64

gone!! ;)

32. anonymous

??

33. anonymous

oh i see what they want

34. anonymous

so is it 1 + 1/p^2 ??

35. anonymous

i don't think so let me draw a triangle. you should draw one too

36. anonymous

|dw:1327848673010:dw|

37. anonymous

there is my triangle, the triangle where tan(x) = p

38. anonymous

now you need the hypotenuse which you find by pythagoras

39. anonymous

|dw:1327848731602:dw|

40. anonymous

hypotenuse is $\sqrt{p^2+1}$ you are looking for $\csc(x)$ and that is $\frac{\sqrt{p^2+1}}{p}$

41. anonymous

nonononon. See, cosec^2x = 1 + cot^2x = 1 + 1/tan^2x = 1 + 1/p^2, right??

42. anonymous

oh sorry you want cosecant squared, so it is the square of what i wrote, which is $\frac{p^2+1}{p^2}=1+\frac{1}{p^2}$ yes you have it!

43. anonymous

i forgot about the square, sorry

44. anonymous

hahahahah, studd. Since you love me, you have to do me a favour. You have to join my group.

45. anonymous

group of what?

46. anonymous

my openstudy group. I created it B) It's called, wait for it, Herp-Derp.