$\frac{1}{{2\cos x\sin x}} - \frac{{\sin x}}{{\cos x}} = \frac{{1 - 2{{\sin }^2}x}}{{2\cos x\sin x}} = \frac{{{{\sin }^2}x + {{\cos }^2}x - 2{{\sin }^2}x}}{{2\cos x\sin x}} = \frac{{{{\cos }^2}x - {{\sin }^2}x}}{{2\cos x\sin x}} = \frac{{\cos 2x}}{{\sin 2x}} = {\mathop{\rm ctg}\nolimits} 2x$