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JayJayDunz · Feb 11, 2009, 10:51 PM

Hi I'm having trouble proving this identity I doing it every which way I can but I'm not getting the sides to match up here is the problem HELP PLZZZZZZ!!

(1-cosC)(1+secC)=tanCsinC

JayJayDunz · Feb 11, 2009, 11:01 PM

Originally Posted by JayJayDunz

Hi I'm havin trouble proving this identity i doing it every which way i can but i'm not gettin the sides to match up here is teh problem HELP PLZZZZZZ!!!!!!

(1-cosC)(1+secC)=tanCsinC

Here's what I did so far working with the right to equal the left... I know its wrong but hey I'm trying here...

tanCsinC= sinC/cosC * sinC = sin2C/cosC = 1-cos2C/cosC = 1/cosC - cos2C/cosC

after that I'm stuck sooooo help please!!

Clough · Feb 11, 2009, 11:09 PM

Hi, JayJayDunz!

Most likely, someone will come along. But, you do need to be patient. Everyone here who answers questions does so voluntarily and on their free time.

The best people to help you right now may be asleep or not logged onto the site.

Since you've posted how you've already tried to solve the problem and not just simply posted the question expecting it to be answered, you'll be more likely to have someone come along who'll be willing to help you.

Thanks!

galactus · Feb 12, 2009, 06:39 AM

expand it out:

$1+sec(C)-cos(C)-cos(C)sec(C)$

But notice that $cos(C)sec(C)=1$

So we end up getting $sec(C)-cos(C)$

$=\frac{1}{cos(C)}-cos(C)$

$=\frac{\overbrace{1-cos^{2}(C)}^{\text{sin^2(C)}}}{cos(C)}$

$=\frac{sin^{2}(C)}{cos(C)}$

$=\underbrace{\frac{sin(C)}{cos(C)}}_{\text{tan(C)} }\cdot sin(C)$

$=tan(C)sin(C)$