proving identity
$\displaystyle \frac {(1 + cosx + sinx)}{( 1 + cosx - sinx)} $=$\displaystyle \frac{ cos x (1 + sin x)} { (1 - sin^2 x)}
$
sorry my mistake.
I assume that you mean this:
$\displaystyle \frac{1 + \cos x + \sin x}{1 - \cos x + \sin x} = \frac{\cos x(1 + \sin x)}{1 - \sin^2 x}$
(Either learn LaTex or use parentheses properly, please!)
First of all, the right side is not simplified:
$\displaystyle \frac{\cos x(1 + \sin x)}{1 - \sin^2 x}$
$\displaystyle \begin{aligned}
&= \frac{\cos x(1 + \sin x)}{\cos^2 x} \\
&= \frac{1 + \sin x}{\cos x}
\end{aligned}$
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.
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