Proving Identities

oryxncrake · #1 $Old$ October 6th, 2009, 03:26 PM

How do I prove the following identity?

cosx (secx - cscx) = 1 - cotx

For the left side, I am at:

cosx (1/cosx - 1/sinx)

Am I anywhere near being right so far? Or is there an easier way to simplify the identity?

pickslides · #2 $Old$ October 6th, 2009, 03:41 PM

Its a really good start

$\cos(x) \left(\frac{1}{\cos(x)} - \frac{1}{\sin(x)}\right)$

now expanding $\cos(x)$ in you get

$\frac{\cos(x)}{\cos(x)} - \frac{\cos(x)}{\sin(x)}$

Spoiler:

oryxncrake · #3 $Old$ October 6th, 2009, 03:52 PM

Quote:

Originally Posted by pickslides $View Post$

Its a really good start

$\cos(x) \left(\frac{1}{\cos(x)} - \frac{1}{\sin(x)}\right)$

now expanding $\cos(x)$ in you get

$\frac{\cos(x)}{\cos(x)} - \frac{\cos(x)}{\sin(x)}$

Spoiler:

Thanks so much. I just get so confused with the simplifying!

meyanne · #4 $Old$ November 4th, 2009, 04:49 AM

hi there

) we are now at that leeson.,and I am proud to say that I'm able to answer your question..and I have an easier way to answer it..here it is:

cos (sec-csc) = 1-cot (start at the left side)
cos sec - cos csc = 1-cot (distribute)
cos(1/cos) - sin/tan(1/sin) = 1-cot (cancel cos and sin)
1-1/tan = 1-cot
1-cot = 1-cot
and that's it!!!

At first, I am also confused with that..but I keep on researching until I learned it!!! hehe..

I hope that I helped you..
O_o