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  #1  
Old November 2nd, 2009, 04:29 PM
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Default Urgent help with verifying trigonmetric identities

1/secxtanx = cscx - sinx

I've tried some ways and every way I do this, I keep ending up in a dead end. It's not only just this problem, there's more I have trouble with. So, if anyone out there has some useful tips and strategies for verifying these trigonometric identities. It would be much appreciated. Thank You.


So anyway, 1/secxtanx = cscx - sinx. Here's what I attempted to do:

1/secxtanx = cscx - sinx
1/(1/cosx)(sinx/cosx) = cscx-sinx
cosx cosx/sinx = cscx-sinx

Here's where I run into trouble. What do I do? Where did I go wrong? I've spent a long time trying to figure this out and I'd love some help right now. So if anyone could help me, thanks.
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Old November 2nd, 2009, 06:12 PM
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Originally Posted by lax600 View Post
1/secxtanx = cscx - sinx

I've tried some ways and every way I do this, I keep ending up in a dead end. It's not only just this problem, there's more I have trouble with. So, if anyone out there has some useful tips and strategies for verifying these trigonometric identities. It would be much appreciated. Thank You.


So anyway, 1/secxtanx = cscx - sinx. Here's what I attempted to do:

1/secxtanx = cscx - sinx
1/(1/cosx)(sinx/cosx) = cscx-sinx
cosx cosx/sinx = cscx-sinx

Here's where I run into trouble. What do I do? Where did I go wrong? I've spent a long time trying to figure this out and I'd love some help right now. So if anyone could help me, thanks.
You're on the right track. You have \frac{\cos^2(x)}{\sin(x)} Use \sin^2(x)+\cos^2(x)=1 and substitute for cos^2(x). You'll get two terms and when you simplify you'll have your answer.
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Old November 2nd, 2009, 06:14 PM
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hint ...

\cos{x}\cdot\cos{x} = \cos^2{x} = 1-\sin^2{x}
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