Chứng minh rằng \(\frac{{2\tan x - \sin 2x}}{{{{\left( {\sin x + \cos x} \right)}^2} - 1}} = {\tan ^2}x\)
Giải chi tiết:
\(\begin{array}{l}VT = \dfrac{{\dfrac{{2\sin x}}{{\cos x}} - 2\sin x\cos x}}{{{{\sin }^2}x + 2\sin x\cos x + {{\cos }^2}x - 1}} \\\,\,\,\,\,\,\,\,= \dfrac{{2\sin x\left( {\dfrac{1}{{\cos x}} - \cos x} \right)}}{{2\sin x\cos x}}\\\;\;\;\;\; = \dfrac{{1 - {{\cos }^2}x}}{{{{\cos }^2}x}} = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} = {\tan ^2}x = VP\;\;\left( {dpcm} \right).\end{array}\)