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tuỳ ý giá trị biểu thức 
Ta có: \(cot\alpha=\dfrac{cos\alpha}{sin\alpha}=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}\)
Lại có: \(\dfrac{1}{cot\alpha}=tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{sin^2\alpha}{cos\alpha.sin\alpha}=\dfrac{1}{\sqrt{5}}\)
\(\Rightarrow A=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}+\dfrac{sin^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}+\dfrac{1}{\sqrt{5}}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)
Ta có : cot α = \(\sqrt{5}\Rightarrow\dfrac{cos\alpha}{sin\alpha}=\sqrt{5}\Rightarrow cos\alpha=\sqrt{5}.sin\alpha\)
\(A=\dfrac{sin^2\alpha+cos^2\alpha}{sin\alpha.cos\alpha}\)
\(A=\dfrac{sin^2\alpha+\left(\sqrt{5}sin\alpha\right)^2}{sin\alpha.\sqrt{5}sin\alpha}=\dfrac{sin^2\alpha+5sin^2\alpha}{\sqrt{5}sin^2\alpha}\)
\(A=\dfrac{6sin^2\alpha}{\sqrt{5}sin^2\alpha}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)
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 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
\(tan^2a=\frac{sin^2a}{cos^2a}=\frac{1-cos^2a}{cos^2a}=\frac{1-\left(\frac{3}{5}\right)^2}{\left(\frac{3}{5}\right)^2}=\frac{16}{9}\Rightarrow\left[{}\begin{matrix}tana=\frac{4}{3}\\tana=-\frac{4}{3}\end{matrix}\right.\)
Với \(tana=\frac{4}{3}\Rightarrow cota=\frac{3}{4}\)
\(A=\frac{\frac{4}{3}+\frac{3}{4}+1}{\frac{4}{3}-\frac{3}{4}+1}=\frac{37}{19}\)
Với \(tana=-\frac{4}{3}\Rightarrow cota=-\frac{3}{4}\)
\(A=\frac{-\frac{4}{3}-\frac{3}{4}+1}{-\frac{4}{3}+\frac{3}{4}+1}=-\frac{13}{5}\)
