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ta có tan a.cot a=1
=>tan a= 1:cot a
thay vào pt ta được 1 : cot a+cot a=3
=> cot a=2,62
ta có \(cos\alpha=\frac{cos\alpha}{sin\alpha}=\frac{131}{50}\)
<=>\(\frac{cosa}{131}=\frac{sina}{50}\)
BP 2 vế :
\(\frac{cos^2a}{131^2}=\frac{sin^2a}{50^2}=\frac{cos^2a+sin^2a}{131^2+50^2}=\frac{1}{19661}\)
=>cos2a=0,873=>cos a=0,934
=>sin2a=0,127=>sin a = 0,356
===>A=sin a.cos a=0,356.0,934=0,332504
Tích nha bạn
\(tan\alpha+cot\alpha=\frac{cos\alpha}{sin\alpha}+\frac{sin\alpha}{cos\alpha}=\frac{cos^2\alpha+sin^2\alpha}{sin\alpha.cos\alpha}=\frac{1}{sin\alpha.cos\alpha}=3\)rồi suy ra sin a .cosa = 1/3
a) ta có : \(sin\alpha.cos\alpha\left(tan\alpha+cot\alpha\right)=sin\alpha.cos\alpha\left(\dfrac{sin\alpha}{cos\alpha}+\dfrac{cos\alpha}{sin\alpha}\right)\)
\(=sin^2\alpha+cos^2\alpha=1\)
b) ta có : \(\left(sin^2\alpha+cos^2\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)
\(=1^2+1-2sin\alpha.cos=2\left(1-2sin\alpha.cos\alpha\right)\)
c) ta có : \(tan^2\alpha-sin^2\alpha.tan^2\alpha=tan^2\alpha\left(1-sin^2\alpha\right)\)
\(=\dfrac{sin^2\alpha}{cos^2\alpha}.cos^2\alpha=sin^2\alpha\)
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/ \(B=\left(cosa-sina\right)^2+\left(cosa+sina\right)^2+cos^4a-sin^4a-2cos^2a\)
\(=cos^2a-2cosa.sina+sin^2a+cos^2a+2cosa.sina+sin^2a+\left(cos^2a+sin^2a\right)\left(cos^2a-sin^2a\right)-2cos^2a\)
\(=2+\left(cos^2a-sin^2a\right)-2cos^2a\)
\(=2-sin^2a-cos^2a=2-1=1\)
\(A=\cot\alpha+\frac{\sin\alpha}{\sin\alpha+\cos\alpha}=\cot\alpha+\frac{1}{1+\cot\alpha}=\frac{1}{\tan\alpha}+\frac{1}{1+\frac{1}{\tan\alpha}}=\frac{1}{2}+\frac{1}{1+\frac{1}{2}}=\frac{7}{6}\)
\(tana+cota=3\Leftrightarrow\frac{sina}{cosa}+\frac{cosa}{sina}=3\)
<=> \(\frac{sin^2a+cos^2a}{sina\cdot cosa}=3\Leftrightarrow\frac{1}{sina.cosa}=3\)
=> sina * cosa = 1/3