\(\dfrac{\left(sina+cosa\right)^2-\left(sina-cosa\right)^2}{sina.cosa}=4\\ VT=\dfrac{sin^2a+2sinacosa+cos^2a-sin^2a+2sinacosa-cos^2a}{sinacosa}\\ =\dfrac{4sinacosa}{sinacosa}=4=VP\)

a: \(S=cos^2a\left(1+tan^2a\right)=cos^2a\cdot\dfrac{1}{cos^2a}=1\)

b: \(VP=\dfrac{1+sin2a-1+sin2a}{\dfrac{1}{2}\cdot sin2a}=\dfrac{2\cdot sin2a}{\dfrac{1}{2}\cdot sin2a}=4=VT\)

A= cos⁴ ∝ + cos² ∝ . sin² ∝ + sin² ∝

=cos⁴ ∝+(cos² ∝+1).sin² ∝

=cos⁴ ∝+(1+cos⁴ ∝)(1-cos⁴ ∝)

=cos⁴ ∝+1-cos⁴ ∝=1

Ta có: \(M=\sin^2B+\cos^2B+\sin^2B+\cos^2B-2\sin B\cos B+2\sin B\cos B=2\left(\sin^2B+\cos^2B\right)=2.1=2\)

a/ \(A=\frac{cot^2a-cos^2a}{cot^2a}-\frac{sina.cosa}{cota}\)

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

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

b/ \(B=\left(cosa-sina\right)^2+\left(cosa+sina\right)^2+cos^4a-sin^4a-2cos^2a\)

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

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

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

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

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

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

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