Ta có:
$$\begin{array}{l} VP = \frac{{\cos x}}{{\sin x}} - \frac{{\cos 16x}}{{\sin 16x}}\\  = \frac{{\cos x}}{{\sin x}} - \frac{{2{{\cos }^2}8x - 1}}{{\sin 16x}}\\  = \frac{{\cos x}}{{\sin x}} - \frac{{2{{\cos }^2}8x}}{{2\cos 8x.\sin 8x}} + \frac{1}{{\sin 16x}}\\  = \frac{{\cos x}}{{\sin x}} - \frac{{\cos 8x}}{{\sin 8x}} + \frac{1}{{\sin 16x}}\\  = \frac{{\cos x}}{{\sin x}} - \frac{{2{{\cos }^2}4x - 1}}{{\sin 8x}} + \frac{1}{{\sin 16x}}\\  = \frac{{\cos x}}{{\sin x}} - \frac{{2{{\cos }^2}4x}}{{2\cos 4x.\sin 4x}} + \frac{1}{{\sin 8x}} + \frac{1}{{\sin 16x}}\\  = \frac{{\cos x}}{{\sin x}} - \frac{{\cos 4x}}{{\sin 4x}} + \frac{1}{{\sin 8x}} + \frac{1}{{\sin 16x}}\\  = \frac{{\cos x}}{{\sin x}} - \frac{{2{{\cos }^2}2x - 1}}{{\sin 4x}} + \frac{1}{{\sin 8x}} + \frac{1}{{\sin 16x}}\\  = \frac{{\cos x}}{{\sin x}} - \frac{{2{{\cos }^2}2x}}{{2\cos 2x.\sin 2x}} + \frac{1}{{\sin 4x}} + \frac{1}{{\sin 8x}} + \frac{1}{{\sin 16x}}\\  = \frac{{\cos x}}{{\sin x}} - \frac{{\cos 2x}}{{\sin 2x}} + \frac{1}{{\sin 4x}} + \frac{1}{{\sin 8x}} + \frac{1}{{\sin 16x}}\\  = \frac{{\cos x}}{{\sin x}} - \frac{{2{{\cos }^2}x - 1}}{{\sin 2x}} + \frac{1}{{\sin 4x}} + \frac{1}{{\sin 8x}} + \frac{1}{{\sin 16x}}\\  = \frac{{\cos x}}{{\sin x}} - \frac{{2{{\cos }^2}x}}{{2\sin x.\cos x}} + \frac{1}{{\sin 2x}} + \frac{1}{{\sin 4x}} + \frac{1}{{\sin 8x}} + \frac{1}{{\sin 16x}}\\  = \frac{{\cos x}}{{\sin x}} - \frac{{\cos x}}{{\sin x}} + \frac{1}{{\sin 2x}} + \frac{1}{{\sin 4x}} + \frac{1}{{\sin 8x}} + \frac{1}{{\sin 16x}} = VT \end{array}$$