anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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
  • anonymous
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
  • anonymous
Sorry

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anonymous
  • anonymous
Made a mistake
Nnesha
  • Nnesha
hmm well....... i guess both sides are not equal
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @Nixy Made a mistake \(\color{blue}{\text{End of Quote}}\) what ?? :-)
anonymous
  • anonymous
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
  • anonymous
I was typing it in too fast and made a boo boo. Sorry about that.
anonymous
  • anonymous
Try sinĀ² x = 1 - cosĀ² x
anonymous
  • anonymous
So the numerator becomes \[1 - cos^2 x - 2 (sin x)(cos x) + cos^2 x\]
Nnesha
  • Nnesha
that's right or just put 1 sin^2x+cos^2 =1 you will get the same answer :-)
anonymous
  • anonymous
That is it peachpi.
anonymous
  • anonymous
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
  • anonymous
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
  • Nnesha
haha great work!!! :-)

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