cosx(1 + 1/cosx + tanx)(1 - 1/cosx + tanx)
= cosx[(1 + tanx)² - 1/cos²x]
= cosx(1 + tan²x + 2tanx - 1 - tan²x)
= 2tanxcosx
= 2sinxcosx/cosx
= 2sinx
Chọn D
\(=cosx\left(1+\dfrac{1}{cosx}+\dfrac{sinx}{cosx}\right)\left(1-\dfrac{1}{cosx}+\dfrac{sinx}{cosx}\right)\)
\(=cosx\left(\dfrac{sinx+1+cosx}{cosx}\right)\left(\dfrac{cosx+sinx-1}{cosx}\right)\)
\(=\left(cosx+sinx\right)^2\)\(-1\)
\(=cos^2x+2cosxsinx+sin^2x-1\)
\(=cos^2x+sin^2x+2cosxsinx-1\)
\(=1+2cosxsinx-1\)
\(=2cosxsinx\)
Áp dụng đẳng thức góc nhân đôi cho sin
\(=2sinx\)