Chứng minh: \(\dfrac{{\sin 4x}}{{1 + \cos 4x}}.\dfrac{{\cos 2x}}{{1 + \cos 2x}} = \tan x\)
Chứng minh: \(\dfrac{{\sin 4x}}{{1 + \cos 4x}}.\dfrac{{\cos 2x}}{{1 + \cos 2x}} = \tan x\)
Ta có \(\dfrac{{\sin 4x}}{{1 + \cos 4x}}.\dfrac{{\cos 2x}}{{1 + \cos 2x}} = \dfrac{{2\sin 2x\cos 2x}}{{2{{\cos }^2}2x}}.\dfrac{{\cos 2x}}{{2{{\cos }^2}x}}\)
\( = \dfrac{1}{2}.\dfrac{{\sin 2x}}{{{{\cos }^2}x}} = \dfrac{1}{2}.\dfrac{{2\sin x\cos x}}{{{{\cos }^2}x}} = \dfrac{{\sin x}}{{\cos x}} = \tan x\) (đpcm)
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