Answer 1

looix

Answer 2

Lại lỗi rồi bạn

Answer 3

lỗi

Answer 4

lx

Answer 5

LỖI

Answer 6

mn giải ch tiết giúp mk với ạ> mk cảm ơn ạ^^

 

Answer 7

Answer 8

\(A=tanx+\dfrac{cosx}{1+sinx}=\dfrac{sinx}{cosx}+\dfrac{cosx}{1+sinx}=\dfrac{sinx\left(1+sinx\right)+cos^2x}{cosx\left(1+sinx\right)}\)

\(=\dfrac{sin^2x+cos^2x+sinx}{cosx\left(1+sinx\right)}=\dfrac{1+sinx}{cosx\left(1+sinx\right)}=\dfrac{1}{cosx}\)

\(B=\dfrac{cosx.tanx}{sin^2x}-cotx.cosx=\dfrac{cosx.\dfrac{sinx}{cosx}}{sin^2x}-\dfrac{cosx}{sinx}.cosx\)

\(=\dfrac{1}{sinx}-\dfrac{cos^2x}{sinx}=\dfrac{1-cos^2x}{sinx}=\dfrac{sin^2x}{sinx}=sinx\)

\(C=\dfrac{1+sinx}{cosx}\left[1-\left(\dfrac{1-sinx}{cosx}\right)^2\right]=\dfrac{1+sinx}{cosx}\left(1-\dfrac{sin^2x-2sinx+1}{cos^2x}\right)\)

\(=\dfrac{1+sinx}{cosx}\left(\dfrac{cos^2x-sin^2x+2sinx-1}{1-sin^2x}\right)=\dfrac{1+sinx}{cosx}\left(\dfrac{cos^2x-sin^2x+2sinx-sin^2x-cos^2x}{\left(1-sinx\right)\left(1+sinx\right)}\right)\)

\(=\dfrac{2sinx-2sin^2x}{cosx\left(1-sinx\right)}=\dfrac{2sinx\left(1-sinx\right)}{cosx\left(1-sinx\right)}=\dfrac{2sinx}{cosx}=2tanx\)