Giả sử các biểu thức đều xác định

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

b/ \(=\frac{sin^2a+\left(1+cosa\right)^2}{sina\left(1+cosa\right)}=\frac{sin^2a+cos^2a+2cosa+1}{sina\left(1+cosa\right)}=\frac{2\left(cosa+1\right)}{sina\left(1+cosa\right)}=\frac{2}{sina}\)

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

Chứng minh các hằng đẳng thức trên

\(\frac{\sin a}{\sin a-cosa}-\frac{cosa}{cosa-\sin a}\)

\(=\frac{\sin a}{\sin a-cosa}+\frac{cosa}{\sin a-cosa}=\frac{\sin a+cosa}{\sin a-cosa}\)

\(\frac{1+\cot^{}a}{1-\cot^{}a}\)

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

\(=\frac{\sin a+cosa}{\sin a}:\frac{\sin a-cosa}{\sin a}=\frac{\sin a+cosa}{\sin a-cosa}\)

Do đó: \(\frac{\sin a}{\sin a-cosa}-\frac{cosa}{cosa-\sin a}=\frac{1+\cot a}{1-\cot a}\)

\(\frac{\sin^2a}{\sin a-cosa}-\frac{\sin a+cosa}{\tan^2a-1}\)

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

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

\(=\frac{\sin^2a}{\sin a-cosa}-\frac{cos^2a}{\sin a-cos^{}a}=\frac{\sin^2a-cos^2a}{\sin a-cosa}=\sin a+cosa\)

\(\dfrac{sina}{sina-cosa}-\dfrac{cosa}{cosa-sina}=\dfrac{sina+cosa}{sina-cosa}=\dfrac{1+cota}{1-cota}=\dfrac{\left(1+cota\right)^2}{1-cot^2a}\)

Đề bài ko đúng

[1-2sina/2cosa/2+(2cos^2a/2 - 1)]/[1-2sina/2cosa/2-1+2sin^a]

=2cosa/2(cosa/2-sina/2)/[2sina/2(sina/2-cosa/2)]

= -cota/2

\(A=\dfrac{cosa+sina}{cosa-sina}=\dfrac{\dfrac{cosa}{cosa}+\dfrac{sina}{cosa}}{\dfrac{cosa}{cosa}-\dfrac{sina}{cosa}}=\dfrac{1+tana}{1-tana}=\dfrac{1+\left(-2\right)}{1-\left(-2\right)}=\dfrac{-1}{3}\)

\(A=\dfrac{1-cosa}{sina}-\dfrac{sina}{1+cosa}=\dfrac{\left(1-cosa\right)\left(1+cosa\right)-sina.sina}{sina\left(1+cosa\right)}\)

\(A=\dfrac{1-cos^2a-sin^2a}{sina\left(1+cosa\right)}=\dfrac{sin^2a-sin^2a}{sina\left(1+cosa\right)}=0\)