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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: \(=\dfrac{cosa+cos^2a+sina}{1+cosa}\)
\(1+tan^2a=1+\frac{sin^2a}{cos^2a}=\frac{cos^2a+sin^2a}{cos^2a}=\frac{1}{cos^2a}\)
\(1+cot^2a=1+\frac{cos^2a}{sin^2a}=\frac{sin^2a+cos^2a}{sin^2a}=\frac{1}{sin^2a}\)
\(cot^2a-cos^2a=\frac{cos^2a}{sin^2a}-cos^2a=cos^2a\left(\frac{1}{sin^2a}-1\right)=cos^2a\left(\frac{1-sin^2a}{sin^2a}\right)\)
\(=cos^2a.\frac{cos^2a}{sin^2a}=cos^2a.cot^2a\)
Câu cuối đề bài sai
b,tan a = sina/cosa
=> \(tan^2a=\frac{sin^2a}{cos^2a}\)
thay vào ta có :
\(cos^2a+\frac{sin^2a}{cos^2a}.cos^2a=cos^2a+sin^2a=1\)
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/ \(cos^2\alpha+tan^2\alpha.cos^2\alpha=cos^2\alpha\left(1+tan^2\alpha\right)\)
\(=cos^2\alpha.\left(1+\frac{sin^2\alpha}{cos^2\alpha}\right)=cos^2\alpha.\left(\frac{sin^2\alpha+cos^2\alpha}{cos^2\alpha}\right)\)
\(=cos^2.\frac{1}{cos^2\alpha}=1\)