\(=\dfrac{2sin2x.cos^22x}{\left(1+2cos^22x-1\right)\left(1+2cos^2x-1\right)}=\dfrac{4sinx.cosx.cos^22x}{4cos^22x.cos^2x}\)
\(=\dfrac{sinx}{cosx}=tanx\)
Đúng 4
Bình luận (1)