Chứng minh rằng \(\frac{{2\tan x - \sin 2x}}{{{{\left( {\sin x + \cos x} \right)}^2} - 1}} = {\tan ^2}x\)

* Hướng dẫn giải

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