Question

[Nguyễn Thảo Hân]

đơn giản biểu thức:

\(\frac{1-cos\alpha}{sin^2\alpha}-\frac{1}{1+cos\alpha}\)

\(\frac{1-sin^2\alpha.cos^2\alpha}{cos^2\alpha}-cos^2\alpha\)

Answer 1

[Nguyễn Việt Lâm]

\(\frac{1-cosa}{1-cos^2a}-\frac{1}{1+cosa}=\frac{1-cosa}{\left(1-cosa\right)\left(1+cosa\right)}-\frac{1}{1+cosa}=\frac{1}{1+cosa}-\frac{1}{1+cosa}=0\)

\(\frac{1-sin^2a.cos^2a}{cos^2a}-cos^2a=\frac{1}{cos^2a}-\frac{sin^2a.cos^2a}{cos^2a}-cos^2a\)

\(=\frac{1}{cos^2a}-\left(sin^2a+cos^2a\right)=\frac{1}{cos^2a}-1\)

\(=\frac{1-cos^2a}{cos^2a}=\frac{sin^2a}{cos^2a}=tan^2a\)