derivate of a trigonometric fraction.
[tex]\frac{1-cosx}{1+cosx}\\\\=\frac{2sin^2\frac{x}{2}}{2Cos^2\frac{x}{2}}\\\\=tan^2\frac{x}{2}\\\\\frac{d}{dx}tan^2\frac{x}{2}=2tan\frac{x}{2}\ *\ sec^2\frac{x}{2}*\frac{1}{2}\\\\another\ way\\\\let\ u=1-cosx,\ \ \ u'=Sinx\\v=1+cosx,\ \ \ v'=-Sinx\\\\\frac{d}{dx}\frac{1-cosx}{1+cosx}=\frac{(1+cosx)(sinx)-(1-cosx)(-sinx)}{(1+cosx)^2}\\\\=\frac{2sinx}{(1+cosx)^2}[/tex]
either way the answer is same.
