có rồi..$1+\frac{1}{cosx}=\frac{2\cos ^{2}\frac{x}{2}}{cosx}$
tương tự rồi nhân lại đk:
$VT=16.\frac{cos^{2}\frac{x}{2}.cosx.cos2x.cos4x}{cos8x}=\frac{cos\frac{x}{2}}{sin\frac{x}{2}}.\frac{16sin\frac{x}{2}.cos\frac{x}{2}.cosx.cos2x.cos4x}{cos8x}$
$=cot\frac{x}{2}.\frac{8sinx.cosx.cos2x.cos4x}{cos8x}=cot\frac{x}{2}.\frac{4sin2x.cos2x.cos4x}{cos8x}$
$=...=cot\frac{x}{2}.\frac{sin8x}{cos8x}=\frac{tan8x}{tan\frac{x}{2}}$
đúng tich hộ mk nha bn....!!!