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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: 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: \(=\dfrac{cosa+cos^2a+sina}{1+cosa}\)
\(\sin^4a.\left(3-2\sin^2a\right)+\cos^4a\left(3-2\cos^2a\right)\)
\(=3\sin^4a-2\sin^6a+3\cos^4a-2\cos^6a\)
\(=3\left(\sin^4a+\cos^4a\right)-2\left(\sin^6a+\cos^6a\right)\)
\(=3\left(\left(\sin^2a\right)^2+\left(\cos^2a\right)^2\right)-2\left(\left(\sin^2a\right)^3+\left(\cos^2a\right)^3\right)\)
\(=3.1-2\left(sin^2a+\cos^2a\right)\left(\sin^4-sin^2.\cos^2+\cos^4\right)\)
\(=3-2.1\left(\left(\sin^2a\right)^2+\left(\cos^2a\right)^2\right).\left(-\sin^2.\cos^2\right)\)
\(=3-2\left(-\sin^2.\cos^2\right)\)
a) \(1+tan^2\alpha=1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2=\dfrac{sin^2\alpha+cos^2\alpha}{cos^2\alpha}=\dfrac{1}{cos^2\alpha}\)
b) \(1+cot^2\alpha=1+\left(\dfrac{cos\alpha}{sin\alpha}\right)^2=\dfrac{cos^2\alpha+sin^2\alpha}{sin^2\alpha}=\dfrac{1}{sin^2\alpha}\)
c) \(tan^2\alpha\left(2sin^2\alpha+3cos^2\alpha-2\right)=tan^2\alpha\left[cos^2\alpha+2\left(sin^2\alpha+cos^2\alpha\right)-2\right]=\dfrac{sin^2\alpha}{cos^2\alpha}\times cos^2\alpha=sin^2\alpha\)
a)
\(1+tan^2\alpha=1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2=\dfrac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}=\dfrac{1}{cos^2\alpha}\)
b)\(1+cot^2\alpha=1+\left(\dfrac{cos\alpha}{sin\alpha}\right)^2=\dfrac{sin^2\alpha+cos^2\alpha}{sin^2\alpha}=\dfrac{1}{sin^2\alpha}\)
c) mình chưa rõ đề nha