Wiskunde Multiple Choice Vraag

$https://latex.codecogs.com/png.image?\dpi{110}D\; \ln\frac{1+\tan \frac x2}{1-\tan \frac x2}=$	A. sin x
B. cos x
C. tan x
D. cot x
E. sec x
F. \(\large\boldsymbol{\frac{2}{\sin2x}}\)

$https://latex.codecogs.com/png.image?\dpi{110}D\; \ln\frac{1+\tan \frac x2}{1-\tan \frac x2}=\frac{1-\tan \frac x2}{1+\tan \frac x2}\cdot D\frac{1+\tan \frac x2}{1-\tan \frac x2}$
$https://latex.codecogs.com/png.image?\dpi{110}\\=\frac{1-\tan \frac x2}{1+\tan \frac x2}\cdot \frac{\frac{1}{2\cos^2\frac x2}(1-\tan \frac x2)-(1+\tan\frac x2)(-\frac 1{2\cos^2\frac x2})}{(1-\tan \frac x2)^2} \\=\frac{1-\tan \frac x2}{1+\tan \frac x2}\cdot \frac{\frac{1}{2\cos^2\frac x2}(1-\tan \frac x2 + 1 + \tan \frac x2)}{(1-\tan \frac x2)^2} \\$ $https://latex.codecogs.com/png.image?\dpi{110}\\=\frac{\frac{1}{\cos^2\frac x2}}{1-\tan^2 \frac x2}=\frac{1}{\cos^2 \frac x2 - \sin \frac x2}=\frac{1}{\cos x}=\sec x$