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Old April 26th, 2009, 03:51 PM
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Default Please simplify

(1+sinx/cosx)-(cosx/sinx-1)

Please simply this
This means no trig function in the denominator.

Thanks!
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Old April 26th, 2009, 05:42 PM
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i can't get the denominator clear i keep getting
-(sinx+1)/cosx or -cosx/(sinx-1)
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Old April 26th, 2009, 06:07 PM
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the final answer would be (1+sinx)-(cosx)

or you can write it as 1+sinx-cosx without the brackets
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Old April 26th, 2009, 06:28 PM
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I'm arriving at a different result...

(1+\frac{sinx}{cosx}) - (\frac{cosx}{sinx}-1)

1 + \frac{sinx}{cosx} - \frac{cosx}{sinx} +1

\frac{sin^2{x}}{sinxcosx} - \frac{cos^2{x}}{sinxcosx}+2

\frac{sin^2{x}-cos^2{x}}{(\frac{1}{2})2sinxcosx} +2

As {cos^2{x}-sin^2{x}} = cos2x and 2sinxcosx=sin2x, we have:

-2\frac{cos2x}{sin2x} +2

\therefore-2cot2x + 2

Last edited by Referos; April 26th, 2009 at 06:28 PM. Reason: latex
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Old April 27th, 2009, 05:32 AM
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Quote:
Originally Posted by decet View Post
(1+sinx/cosx)-(cosx/sinx-1)

Please simply this
This means no trig function in the denominator.

Thanks!
Please post your expressions more clearly. Do you mean \frac{1 + \sin x}{\cos x} - \frac{\cos x}{\sin x - 1} ?
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