precalculus?!, i would love some help, please. |
 Nov 29 2007, 10:17 PM
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#1
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 mediocre one  Group: Member Posts: 7 Joined: Sep 2004 Member No: 51,556  |
Verify that secX(tanX+cotX)=cscX/cos^2X
I've been bonking my head all day!! If someone could help me with this I'd greatly appreciate it. |
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Nov 29 2007, 10:20 PM
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#2
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 Resource Center Tyrant Group: Official Member Posts: 2,263 Joined: Nov 2007 Member No: 593,306 |
I haven't done this in forever. Bear with me.
Work with the left side (or right because both works, but for my purposes, I'm just going to use the left), and turn secX, tanX, and cotX into terms you already know. We know that secX = 1/cosX, tanX=sinX/cosX, and cotX = cosX/sinX If you don't, you must remember this and not blow it off because it will come back to haunt you. Treat it like it's your multiplication table again and get that hammered into your head. 1/cosX(sinX/cosX+cosX/sinX) Multiply: sinX/cos^2X + 1/sinX And it turns into: sin^2X+cos^2X/cos^2X*sinX = 1/cos^2X*sinX 1/cos^2X*1/sinX 1/cosX*csc/1 = cscX/cos^2X |
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Nov 29 2007, 10:29 PM
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#3
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mediocre one Group: Member Posts: 7 Joined: Sep 2004 Member No: 51,556 |
you did that so quickly... are you sure it's right??
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Nov 29 2007, 10:31 PM
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#4
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Resource Center Tyrant Group: Official Member Posts: 2,263 Joined: Nov 2007 Member No: 593,306 |
These trigonometric identities? I can verify any of them in a matter of minutes. I might have been wrong because I typed it rather quickly. Is there something you find that's incorrect?
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Nov 29 2007, 10:48 PM
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#5
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mediocre one Group: Member Posts: 7 Joined: Sep 2004 Member No: 51,556 |
Okay.. thanks. You are a genius!
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Nov 29 2007, 10:52 PM
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#6
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Resource Center Tyrant Group: Official Member Posts: 2,263 Joined: Nov 2007 Member No: 593,306 |
Not really. If your algebra skills are strong, this comes easily.
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