Dani M.

# Prove that these are identities

2cosx - secx = cosx - tanx/cscx

Michael J.

Use parentheses.
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03/28/17

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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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