Trig / Pre Cal/identities Am I on the right path?

Please do not give the answer. Here is the problem

$ \frac{(\sin x- \cos x)*2}{cos} = \sec x - \cos x$

I am working on the left side and need to prove it equals the right. This is what I have done so far

$ \frac{(\sin^2 x - 2(\cos x)(\sin x)+\cos^2 x}{\cos x} = \sec x - \cos x$

Am I going about this the right way because I don't see where to go from here. Thank you.

Question

Trig / Pre Cal/identities Am I on the right path?

Please do not give the answer.  Here is the problem

$   \frac{(\sin x- \cos x)*2}{cos} = \sec x - \cos x$

I am working on the left side and need to prove it equals the right. This is what I have done so far

$   \frac{(\sin^2 x - 2(\cos x)(\sin x)+\cos^2 x}{\cos x} = \sec x - \cos x$

Am I going about this the right way because I don't see where to go from here. Thank you.

anonymous · Answer

It is suppose to be 

Trig / Pre Cal/identities Am I on the right path?

Please do not give the answer.  Here is the problem

$   \frac{(\sin x- \cos x)^2}{cos} = \sec x - \cos x$

I am working on the left side and need to prove it equals the right. This is what I have done so far

$   \frac{(\sin^2 x - 2(\cos x)(\sin x)+\cos^2 x}{\cos x} = \sec x - \cos x$

Am I going about this the right way because I don't see where to go from here. Thank you.

anonymous · Answer

This is where I am at. Am I on the right path and where do I go from here because I do not see anything.

$   \frac{\sin^2 x - 2(\cos x)(\sin x)+\cos^2 x}{\cos x} = \sec x - \cos x$

anonymous · Answer

Sorry

anonymous · Answer

Made a mistake

nnesha · Answer

hmm well.......
i guess both sides are not equal

nnesha · Answer

$\color{blue}{\text{Originally Posted by}}$ @Nixy
Made a mistake
$\color{blue}{\text{End of Quote}}$
what ?? :-)

anonymous · Answer

It is suppose to be 

$ \frac{(\sin x- \cos x)^2}{cos} = \sec x - 2\sin x $

And I am at 
$  \frac{\sin^2 x - 2(\cos x)(\sin x)+\cos^2 x}{\cos x} = \sec x - 2\sin x  $

anonymous · Answer

I was typing it in too fast and made a boo boo. Sorry about that.

anonymous · Answer

Try sin² x = 1 - cos² x

anonymous · Answer

So the numerator becomes 
$$1 - cos^2 x - 2 (sin x)(cos x) + cos^2 x$$

nnesha · Answer

that's right or just put 1 
sin^2x+cos^2 =1 
you will get the same answer :-)

anonymous · Answer

That is it peachpi.

anonymous · Answer

I did that a min ago and it did not work but re looked at it and I see where I made my mistake. Thank you all

anonymous · Answer

I think I over think these things at time. One minute I am rolling through them and then I get snagged by the simplest thing lol Thank you all

nnesha · Answer

haha great work!!! :-)