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?

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- Oct 6th 2009, 01:26 PMoryxncrakeProving Identities
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? - Oct 6th 2009, 01:41 PMpickslides
Its a really good start

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

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

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

__Spoiler__: - Oct 6th 2009, 01:52 PMoryxncrake
- Nov 4th 2009, 02:49 AMmeyanneeasier answer O_o
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