Show: $\displaystyle \frac{\sin^2x}{1-\cos x} = \frac{1 + \sec}{\sec x}$
Thanks for your help.
Hi Savior_Self,
I'm going to mess with both sides in order to prove this identity.
$\displaystyle \dfrac{\sin^2x}{1-\cos x}=\dfrac{1+\sec x}{\sec x}$
$\displaystyle \dfrac{1-\cos^2 x}{1-\cos x}=\dfrac{1+\dfrac{1}{\cos x}}{\dfrac{1}{\cos x}}$
$\displaystyle \dfrac{(1-\cos x)(1+\cos x)}{1-\cos x}=\dfrac{\dfrac{\cos x+1}{\cos x}}{\dfrac{1}{\cos x}}$
$\displaystyle \cos x+1=\cos x+1$
Moo hath beaten me to the punch!
Alright guys.. I asked for help with this problem a few weeks ago, and I just received the graded worksheet. I liked what you guys did with this, I thought it was clever. But, apparently, this wasn't the answer the lady was looking for because it's marked completely wrong. I'm about to start trying to figure it out myself, but I can always use some help from you guys. So my question is: Is there any way to work through this using only the left side? Starting with $\displaystyle \frac{\sin^2 x}{1 - \cos x}$ and, changing only that, showing that it equals $\displaystyle \frac{1 + \sec x}{\sec x}$?
Thanks! and sorry to ask again.
Hi Savior_Self,
Yes, there is and your instructor should have given you better instructions about which side she wanted to work with. Here is just the left side manipulated:
$\displaystyle \frac{\sin^2 x}{1-\cos x}=$
$\displaystyle \frac{1-\cos^2 x}{1-\cos x}=$
$\displaystyle \frac{(1-\cos x)(1+\cos x)}{1-\cos x}=$
$\displaystyle 1+\cos x =$
$\displaystyle 1+\frac{1}{\sec x}=$
$\displaystyle \frac{\sec x + 1}{\sec x}=\frac{1+\sec x}{\sec x}$