Chứng minh các đẳng thức:a) \(\frac{{\sin 6x.\cos 4x - \cos 6x\sin 4x}}{{1 + \cos 2a}} = \tan x\)b) \(\cot x.

* Hướng dẫn giải

a) \(\frac{{\sin 6x.\cos 4x - \cos 6x\sin 4x}}{{1 + \cos 2a}} = \frac{{\sin 2x}}{{1 + \cos 2x}} = \frac{{2\sin x\cos x}}{{2{{\cos }^2}x}} = \frac{{\sin x}}{{\cos x}} = \tan x\)

b) 

\(\begin{array}{l}
\cot x.\cos \left( {\pi  - x} \right) + \frac{{\sin \left( {\frac{\pi }{2} - x} \right)\tan x}}{{1 - {{\cos }^2}x}}\\
 =  - \cot x.\cos x + \frac{{\cos x.\tan x}}{{{{\sin }^2}x}} = \frac{{ - {{\cos }^2}x}}{{\sin x}} + \frac{1}{{\sin x}}\\
 = \frac{{1 - {{\cos }^2}x}}{{\sin x}} = \frac{{{{\sin }^2}x}}{{\sin x}} = \sin x
\end{array}\)