precalculus?!, i would love some help, please. |
 Nov 29 2007, 10:17 PM
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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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Posts in this topic
rebecco precalculus?! Nov 29 2007, 10:17 PM
rebecco you did that so quickly... are you sure it's r... Nov 29 2007, 10:29 PM MissHygienic These trigonometric identities? I can verify any o... Nov 29 2007, 10:31 PM
rebecco Okay.. thanks. You are a genius! Nov 29 2007, 10:48 PM MissHygienic Not really. If your algebra skills are strong, thi... Nov 29 2007, 10:52 PM |
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