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

Đề bài ko đúng

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

Giải:

\(VP=\frac{sina+sin2a}{1+cosa+cos2a}=\frac{sina+2sinacosa}{1+cosa+2cos^2a-1}=\frac{sina\left(1+2cosa\right)}{cosa\left(1+2cosa\right)}=\frac{sina}{cosa}=tana=VT\)

=> ĐPCM

a.

\(\dfrac{sina+sin5a+sin3a}{cosa+cos5a+cos3a}=\dfrac{2sin3a.cosa+sin3a}{2cos3a.cosa+cos3a}=\dfrac{sin3a\left(2cosa+1\right)}{cos3a\left(2cosa+1\right)}=\dfrac{sin3a}{cos3a}=tan3a\)

b.

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

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

\(sina+cosa=\dfrac{1}{2}\Rightarrow\left(sina+cosa\right)^2=\dfrac{1}{4}\Rightarrow2sinacosa=\dfrac{1}{4}-1=\dfrac{-3}{4}\)

\(\Leftrightarrow-2sinacosa=\dfrac{3}{4}\)

\(\Leftrightarrow cos^2a+sin^2a-2sinacosa=cos^2a+sin^2a+\dfrac{3}{4}\)

\(\Rightarrow\left(sina-cosa\right)^2=1+\dfrac{3}{4}=\dfrac{7}{4}\)

\(\Rightarrow\left|sina-cosa\right|=\dfrac{\sqrt{7}}{2}\)

[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