\(A=\left(sin^2a+cos^2a\right)^3-3\cdot sin^2a\cdot cos^2a\left(sin^2a+cos^2a\right)+3\cdot sin^2a\cdot cos^2a\)

\(=1-3\cdot sin^2a\cdot cos^2a+3\cdot sin^2a\cdot cos^2a\)

=1

a, A = 2

b, B = 1

b: \(=\left(\cos^2\alpha+\sin^2\alpha\right)^3-3\cos^2\alpha\sin^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)+3\cdot\sin^2\alpha\cdot\cos^2\alpha\)

=1

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

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

\(=2cos^2a\)

\(cos^6a+sin^6a+3sin^2a.cos^2a\)

\(=\left(cos^2a+sin^2a\right)^3-3sin^2a.cos^2a\left(sin^2a+cos^2a\right)+3sin^2a.cos^2a\)

\(=1-3sin^2a.cos^2a.1+3sin^2a.cos^2a\)

\(=1\)

$\sin^4 a-cos^4 a+2\sin^2 a.\cos^2 a\\=(\sin^4 a-\cos^4 a)+2\sin^2 a.\cos^2 a\\=(\sin^2 a+\cos^2 a)(\sin^2-\cos ^2 )+2\sin^2 a.\cos^2 a\\=\sin^2 a-\cos^2 a+2\sin^2 a.\cos^2 a$

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

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

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

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

A= \(\left(\sin^2a\right)^3+\left(cos^2a\right)^3+3sin^2acos^2a\)

=\(\left(sin^2a+cos^2a\right)\left(sin^4a-cos^2asin^2a+cos^4a\right)+3sin^2acos^2a\)

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

( tan²a+cot a)^{2 _}  ( tan a - cot a )²

\(\left(1+tan^2a\right)\left(1-sin^2a\right)-\left(1+cot^2a\right)\left(1-cos^2a\right)\)

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

\(=cos^2a+sin^2a-sin^2a-cos^2a=\)\(0\)

Vậy B=0