Prove that

1-sin^2X/1+cotX - cos^2X/1+tanX = sinXcosX

Much thanks in advance. (Nod)

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- July 28th 2008, 02:35 PMahlingTrig Identities
Prove that

1-sin^2X/1+cotX - cos^2X/1+tanX = sinXcosX

Much thanks in advance. (Nod) - July 28th 2008, 05:49 PMTKHunny
- July 28th 2008, 06:16 PMahling
Yes, I understand, but .. in the original equation, there are no parantheses. I typed the question exactly as it is shown on the paper given to me.

- July 28th 2008, 07:23 PMSoroban
Hello, ahling!

Quote:

Prove that: .

Multiply the first fraction by , the second by

. .

. .

. .

. . . . . ta-*DAA!*

And LateX is working !!

- July 29th 2008, 06:24 AMTKHunny
I have a little trouble believing that, but let's go with it...

1-sin^2X/1+cotX - cos^2X/1+tanX = sinXcosX

The "/1" are meaningless in this form, so...

1-sin^2X+cotX - cos^2X+tanX = sinXcosX

**1-sin^2X = cos^2X <== Identity to Substitute**

**cos^2X**+cotX - cos^2X+tanX = sinXcosX

cotX+tanX = sinXcosX

Since cotX = (cosX)/(sinX) and tanX = (sinX)/(cosX)

And Since cos^2X + sin^2X = 1

1/(sinXcosX) = sinXcosX

I think you would have a hard time selling that as an identity.

Note: I would do this on an exam. It is technically correct, but may have points taken off for being a smart aleck. (Evilgrin) I would also rewrite it as it probably is intended and work it that way. This may avoid the smart aleck points. Don't put up with sloppy notation. Certainly don't write any yourself. Everyone needs reminders to be clear - particularly if they want to be understood generally.

LaTeX Check: