Question

[Nguyễn Thị Bích Ngọc]

đơn giản biểu thức:tanx + (cosx/(1+sinx))

Giup mik vs

Answer 1

[nguyễn thi nga]

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

Answer 2

[KAl(SO4)2·12H2O]

\(tanx+\frac{cosx}{1+sinx}\)

\(=\frac{cosx}{1+sinx}+\frac{sinx}{cosx}\)

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

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

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

\(=\frac{sinx+1}{cosx.\left(sinx+1\right)}\)

\(=\frac{1}{cosx}\)