identity

**z1llch** · July 20th 2009, 02:58 AM

proving identity
$\frac {(1 + cosx + sinx)}{( 1 + cosx - sinx)}$ = $\frac{ cos x (1 + sin x)} { (1 - sin^2 x)}<br />$
sorry my mistake.

**yeongil** · July 20th 2009, 03:21 AM

I assume that you mean this:
$\frac{1 + \cos x + \sin x}{1 - \cos x + \sin x} = \frac{\cos x(1 + \sin x)}{1 - \sin^2 x}$

(Either learn LaTex or use parentheses properly, please!)

First of all, the right side is not simplified:
$\frac{\cos x(1 + \sin x)}{1 - \sin^2 x}$
$\begin{aligned}<br /> &= \frac{\cos x(1 + \sin x)}{\cos^2 x} \\<br /> &= \frac{1 + \sin x}{\cos x}<br /> \end{aligned}$

If you do this, then this problem is the exact same one I solved earlier for arze in this thread: http://www.mathhelpforum.com/math-he...255-prove.html.

01

**z1llch** · July 20th 2009, 03:27 AM

did tried to use conjugate. but couldnt simplify it..

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