tôi có lớp 6 thôi ạ cái này lớp 9 thì bó tay thôi 

chị đã ghi rõ là toán lớp 9 mà em ko biết làm cũng phải thôi :))

\(sina=\frac{3}{5}\Rightarrow sin^2a=\frac{9}{25}\) ; \(cos^2a=1-\frac{9}{25}=\frac{16}{25}\)

\(A=\frac{cota+tana}{cota-tana}=\frac{sina.cosa\left(cota+tana\right)}{sina.cosa\left(cota-tana\right)}=\frac{cos^2a+sin^2a}{cos^2a-sin^2a}=\frac{1}{cos^2a-sin^2a}=\frac{1}{\frac{16}{25}-\frac{9}{25}}=\frac{25}{7}\)

\(B=\frac{sin^2a-cos^2a}{sin^2a-3cos^2a}=\frac{\frac{sin^2a}{sin^2a}-\frac{cos^2a}{sin^2a}}{\frac{sin^2a}{sin^2a}-\frac{3cos^2a}{sin^2a}}=\frac{1-cot^2a}{1-3cot^2a}=\frac{1-\left(-\frac{1}{3}\right)^2}{1-3\left(-\frac{1}{3}\right)^2}=\)

\(C_1=sin^2a+cos^2a+cos^2a=1+cos^2a=1+\frac{1}{1+tan^2a}=1+\frac{1}{1+\left(-2\right)^2}\)

\(C_2=\left(sin^2a+cos^2a\right)\left(sin^2a-cos^2a\right)=sin^2a-cos^2a=1-2cos^2a\)

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

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

b) \(\cot^2\alpha+1=\frac{\cos^2\alpha}{\sin^2\alpha}+1=\frac{\cos^2\alpha+\sin^2\alpha}{\sin^2\alpha}=\frac{1}{\sin^2\alpha}\)

c) \(\cos^4\alpha-\sin^4\alpha=\left(\cos^2\alpha+\sin^2\alpha\right)\left(\cos^2\alpha-\sin^2\alpha\right)=\cos^2\alpha-\sin^2\alpha\)

\(=2\cos^2\alpha-\left(\sin^2\alpha+\cos^2\alpha\right)=2\cos^2-1\)

\(\cos^4\alpha-\sin^4\alpha+1\\ =\left(\sin^2\alpha+\cos^2\alpha\right)\left(-\sin^2\alpha+\cos^2\alpha\right)+\left(\sin^2\alpha+\cos^2\alpha\right)\\ =-\sin^2\alpha+\cos^2\alpha+\sin^2\alpha+\cos^2\alpha=2\cos^2\alpha\)

\(cos^4a-sin^4a+1=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1\)

\(=cos^2a-sin^2a+1=cos^2a-sin^2a+sin^2a+cos^2a=2cos^2a\)

Vậy ta có đpcm 

\(sin^2a+cos^2a-sin^4a-2cos^2a+sin^2a\)

\(=2sin^2a-cos^2a-sin^4a\)

\(=2sin^2a-cos^2a-\left(\frac{1-cos2a}{2}\right)^2\)

khai triển ra rồi quy đồng lên

\(=\frac{8sin^2a-4cos^2a-1+2cos2a-cos^22a}{4}\)

Mà \(2cos2a=2\left(cos^2a-1\right)=4cos^2-2\)

\(\Rightarrow\frac{8sin^2a-cos^22a-3}{4}\)

Mà \(-cos^22a=sin^22a-1=4sin^2cos^2-1\)

\(\Rightarrow\frac{8sin^2a+4sin^2a.cos^2a-4}{4}\)

\(=\frac{4sin^2a.\left(2-cos^2a\right)-4}{4}\)

\(=sin^2a\left(1+sin^2a\right)-1\)

\(=sin^4a-cos^2a\)

viết lại đề đi cậu ơi