1. ## identity problem

Show: $\frac{\sin^2x}{1-\cos x} = \frac{1 + \sec}{\sec x}$

2. Hello,

$\frac{\sin^2x}{1-\cos x}=\frac{1-\cos^2x}{1-\cos x}=1+\cos x$

and $\frac{1+\sec x}{\sec x}=\frac{1}{\sec x}+1=\cos x+1$

3. Originally Posted by Savior_Self
Show: $\dfrac{\sin^2x}{1-\cos x} = \dfrac{1 + \sec x}{\sec x}$

Hi Savior_Self,

I'm going to mess with both sides in order to prove this identity.

$\dfrac{\sin^2x}{1-\cos x}=\dfrac{1+\sec x}{\sec x}$

$\dfrac{1-\cos^2 x}{1-\cos x}=\dfrac{1+\dfrac{1}{\cos x}}{\dfrac{1}{\cos x}}$

$\dfrac{(1-\cos x)(1+\cos x)}{1-\cos x}=\dfrac{\dfrac{\cos x+1}{\cos x}}{\dfrac{1}{\cos x}}$

$\cos x+1=\cos x+1$

Moo hath beaten me to the punch!

4. ahhh, thank you. Your reply made it easier to understand Moo's, masters. So worry not. You both contributed much to my instruction. haha, thanks again.

5. 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 $\frac{\sin^2 x}{1 - \cos x}$ and, changing only that, showing that it equals $\frac{1 + \sec x}{\sec x}$?

Thanks! and sorry to ask again.

6. Originally Posted by Savior_Self
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 $\frac{\sin^2 x}{1 - \cos x}$ and, changing only that, showing that it equals $\frac{1 + \sec x}{\sec x}$?

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:

$\frac{\sin^2 x}{1-\cos x}=$

$\frac{1-\cos^2 x}{1-\cos x}=$

$\frac{(1-\cos x)(1+\cos x)}{1-\cos x}=$

$1+\cos x =$

$1+\frac{1}{\sec x}=$

$\frac{\sec x + 1}{\sec x}=\frac{1+\sec x}{\sec x}$

7. Originally Posted by masters
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:

$\frac{\sin^2 x}{1-\cos x}=$

$\frac{1-\cos^2 x}{1-\cos x}=$

$\frac{(1-\cos x)(1+\cos x)}{1-\cos x}=$

$1+\cos x =$

$1+\frac{1}{\sec x}=$

$\frac{\sec x + 1}{\sec x}=\frac{1+\sec x}{\sec x}$
May I ask how you manipulated $1 + \frac{1}{\sec x}$ into $\frac{\sec x + 1}{\sec x}$?

8. Originally Posted by Savior_Self
May I ask how you manipulated $1 + \frac{1}{\sec x}$ into $\frac{\sec x + 1}{\sec x}$?
$1=\frac{\sec x}{\sec x}$, so

$1+\frac{1}{\sec x}=\frac{\sec x}{\sec x}+\frac{1}{\sec x}=\frac{\sec x+1}{\sec x}$

9. Originally Posted by masters
$1=\frac{\sec x}{\sec x}$, so

$1+\frac{1}{\sec x}=\frac{\sec x}{\sec x}+\frac{1}{\sec x}=\frac{\sec x+1}{\sec x}$
Ahh, I'm an idiot. I'm so worried about trig identities and whatnot I forget basic rational addition.

But thanks, masters. I really appreciate it.