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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/ \(B=\left(cosa-sina\right)^2+\left(cosa+sina\right)^2+cos^4a-sin^4a-2cos^2a\)
\(=cos^2a-2cosa.sina+sin^2a+cos^2a+2cosa.sina+sin^2a+\left(cos^2a+sin^2a\right)\left(cos^2a-sin^2a\right)-2cos^2a\)
\(=2+\left(cos^2a-sin^2a\right)-2cos^2a\)
\(=2-sin^2a-cos^2a=2-1=1\)
Ta có:
\(sin^4a+cos^4a=\left(sin^2a+cos^2a\right)^2-2sin^2acos^2a=1-2sin^2a.cos^2a\)
Và:
\(sin^6a+cos^6a=\left(sin^2a+cos^2a\right)^3-3sin^2a.cos^2a.\left(sin^2a+cos^2a\right)\)
\(=1-3sin^2a.cos^2a\)
Do đó:
\(A=3\left(1-2sin^2a.cos^2a\right)-2\left(1-3sin^2a.cos^2a\right)=1\)
\(B=1-3sin^2.cos^2a+3sin^2a.cos^2a=1\)
\(A=\sin^6\alpha+cos^6\alpha+3\sin^2\alpha\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right).\)vì\(\sin^2\alpha+\cos^2\alpha=1\)
\(=\left(\sin^2\alpha+\cos^2\alpha\right)^3=1^3=1\)
\(B=2\left(\cos^2\alpha+\sin^2\alpha\right)=2.1=2\)
\(C=\frac{-4\cos\alpha\sin\alpha}{\sin\alpha\cos\alpha}=-4\)
Bài 2:
a: \(\sin a=\sqrt{1-\left(\dfrac{4}{5}\right)^2}=\dfrac{3}{5}\)
\(P=4\cdot\sin^2a-6\cdot\cos^2a\)
\(=4\cdot\dfrac{9}{25}-6\cdot\dfrac{16}{25}\)
\(=\dfrac{36-64}{25}=\dfrac{-28}{25}\)
b: \(A=\sin^6a+\cos^6a+3\cdot\sin^2a\cdot\cos^2a\)
\(=\left(\sin^2a+\cos^2a\right)^3-3\sin^2a\cdot\cos^2a\cdot\left(\sin^2a+\cos^2a\right)+3\cdot\sin^2a\cdot\cos^2a\)
\(=1-3\sin^2a\cdot\cos^2a+3\sin^2a\cdot\cos^2a\)
=1