• Apr 26th 2009, 03:51 PM
decet
(1+sinx/cosx)-(cosx/sinx-1)

This means no trig function in the denominator.

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

or you can write it as 1+sinx-cosx without the brackets
• Apr 26th 2009, 06:28 PM
Referos
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:

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

$\therefore-2cot2x + 2$
• Apr 27th 2009, 05:32 AM
mr fantastic
Quote:

Originally Posted by decet
(1+sinx/cosx)-(cosx/sinx-1)

This means no trig function in the denominator.

Thanks!

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