\(\large D\; \ln\frac{1+\tan \frac x2}{1-\tan \frac x2}=\frac{1-\tan \frac x2}{1+\tan \frac x2}\cdot D\frac{1+\tan \frac x2}{1-\tan \frac x2}\)
\(\large=\frac{1-\tan \frac x2}{1+\tan \frac x2}\cdot \frac{\frac{1}{2\cos^2\frac x2}(1-\tan \frac x2)-(1+\tan\frac x2)(-\frac 1{2\cos^2\frac x2})}{(1-\tan \frac x2)^2} \\=\frac{1-\tan \frac x2}{1+\tan \frac x2}\cdot \frac{\frac{1}{2\cos^2\frac x2}(1-\tan \frac x2 + 1 + \tan \frac x2)}{(1-\tan \frac x2)^2}\)
\(\large=\frac{\frac{1}{\cos^2\frac x2}}{1-\tan^2 \frac x2}=\frac{1}{\cos^2 \frac x2 - \sin \frac x2}=\frac{1}{\cos x}=\sec x \)