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a/\(sin^4\alpha+cos^4\alpha+2sin^2\alpha.cos^2\alpha=\left(sin^2\alpha+cos^2\alpha\right)^2=1\)
b/ \(tan^2\alpha-sin^2\alpha.tan^2\alpha=tan^2\alpha\left(1-sin^2\alpha\right)=\frac{sin^2\alpha}{cos^2\alpha}.cos^2\alpha=sin^2\alpha\)
c/ \(cos^2\alpha+tan^2\alpha.cos^2\alpha=cos^2\alpha\left(1+tan^2\alpha\right)\)
\(=cos^2\alpha.\left(1+\frac{sin^2\alpha}{cos^2\alpha}\right)=cos^2\alpha.\left(\frac{sin^2\alpha+cos^2\alpha}{cos^2\alpha}\right)\)
\(=cos^2.\frac{1}{cos^2\alpha}=1\)
a/ \(\left(1-cos\alpha\right)\left(1+cos\alpha\right)=1-cos^2\alpha=\left(sin^2\alpha+cos^2\alpha\right)-cos^2\alpha=sin^2\alpha\)
b/ \(1+sin^2\alpha+cos^2\alpha=1+1=2\)
c/ \(sin\alpha-sin\alpha.cos^2\alpha=sin\alpha\left(1-cos^2\alpha\right)=sin\alpha.sin^2\alpha=sin^3\alpha\)
a: Sửa đề: \(A=sin^2a+sin^2a\cdot tan^2a\)
\(=sin^2a\left(1+tan^2a\right)=sin^2a\cdot\dfrac{1}{cos^2a}=tan^2a\)
b: \(=\dfrac{\left(sina+cosa\right)^2}{sina+cosa}-cosa=sina+cosa-cosa=sina\)
c: \(=\dfrac{cosa+cos^2a+sina}{1+cosa}\)
a)ta có cos2a = 1-sin2a => A = 4(1-sin2a) -6sin2a
A= 4 -10sin2a = 4- 10.(4/5)2 = -2,4
A = -2,4
b) B = tt
