Question

[nguyen thi nhu y]

c/m :\(\dfrac{tanx-sinx}{sin^3x}=\dfrac{1}{cosx\left(1+cosx\right)}\)

Answer 1

[Bùi Thị Vân]

Giả sử biểu thức xác định:
\(\dfrac{tanx-sinx}{sin^3x}=\dfrac{\dfrac{sinx}{cosx}-sinx}{sin^3x}=\dfrac{sinx-cosxsinx}{cosxsin^3x}\)
\(=\dfrac{sinx\left(1-cosx\right)}{sin^3xcosx}\)\(=\dfrac{1-cosx}{cosxsin^2x}=\dfrac{1-cosx}{cosx\left(1-cos^2x\right)}=\dfrac{1}{cosx\left(1+cosx\right)}\).

Answer 2

[nguyen thi nhu y]

cam on nhieu a