Dani M.
asked • 03/28/17Prove that these are identities
2cosx - secx = cosx - tanx/cscx
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1 Expert Answer
Arturo O. answered • 03/28/17
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2cosx - secx = 2cosx - 1/cosx = (2cos2x - 1)/cosx
= [2cos2x - (cos2x + sin2x)]/cosx
= (cos2x - sin2x)/cosx = cosx - sinx(sinx/cosx) = cosx - tanx/cscx
Dani M.
Thank you for answering! But I am confused on something, how did you get (2cos²x-1)/cosx ?
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03/30/17
Arturo O.
Do you see how I got to
2cosx - 1/cosx ?
I want to combine these 2 terms, so I multiplied and divided the first term by cosx to get a common denominator:
2cosx - 1/cosx = 2cosx(cosx/cos) - 1/cosx = [2cosx(cosx) - ] / cosx = (2cos2x - 1)/cos x
Do you see it now?
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03/30/17
Arturo O.
The editing above came out poorly. Here it is again:
2cosx - 1/cosx = 2cosx(cosx/cosx) - 1/cosx = [2cosx(cosx) - 1] / cosx = (2cos2x - 1)/cos x
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03/30/17
Dani M.
Oh okay that makes sense! Thank you so much for explaining it! Much appreciated!
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03/30/17
Arturo O.
You are welcome, Dani.
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03/30/17
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