Question

rút gọn biểu thức :
A = 1 + \(\dfrac{2\sin\alpha.\cos\alpha}{\cos^2\alpha-\sin^2\alpha}\)
B = \(\sin^6\alpha+\cos^6\alpha+3\sin^2\alpha.\cos^2\alpha\)

Answer 1

sữa đề chút nha :

+) ta có : \(A=\dfrac{1+2sin\alpha.cos\alpha}{cos^2\alpha-sin^2\alpha}=\dfrac{\left(sin\alpha+cos\alpha\right)^2}{\left(sin\alpha+cos\alpha\right)\left(cos\alpha-sin\alpha\right)}=\dfrac{sin\alpha+cos\alpha}{cos\alpha-sin\alpha}\)

+) ta có :

\(B=sin^6\alpha+cos^6\alpha+3sin^2\alpha.cos^2\alpha\)

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

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