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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 = \(\sqrt{5}\Rightarrow\dfrac{cos\alpha}{sin\alpha}=\sqrt{5}\Rightarrow cos\alpha=\sqrt{5}.sin\alpha\)
\(A=\dfrac{sin^2\alpha+cos^2\alpha}{sin\alpha.cos\alpha}\)
\(A=\dfrac{sin^2\alpha+\left(\sqrt{5}sin\alpha\right)^2}{sin\alpha.\sqrt{5}sin\alpha}=\dfrac{sin^2\alpha+5sin^2\alpha}{\sqrt{5}sin^2\alpha}\)
\(A=\dfrac{6sin^2\alpha}{\sqrt{5}sin^2\alpha}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)
tuỳ ý giá trị biểu thức
b) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Leftrightarrow\cos^2\alpha=\dfrac{16}{25}\)
hay \(\cos\alpha=\dfrac{4}{5}\)
Ta có: \(A=5\cdot\sin^2\alpha+6\cdot\cos^2\alpha\)
\(=5\cdot\left(\dfrac{3}{5}\right)^2+6\cdot\left(\dfrac{4}{5}\right)^2\)
\(=5\cdot\dfrac{9}{25}+6\cdot\dfrac{16}{25}\)
\(=\dfrac{141}{25}\)
c) Ta có: \(\tan\alpha=\dfrac{1}{\cot\alpha}=\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}\)
\(D=\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\)
\(=\dfrac{\dfrac{9}{16}+\dfrac{16}{9}}{\dfrac{9}{16}-\dfrac{16}{9}}=-\dfrac{337}{175}\)
Ta có :
* sin2α + cos2α = 1
\(\Leftrightarrow\dfrac{3}{4}\) + cos2α = 1
\(\Leftrightarrow\) cos2α \(=\dfrac{1}{4}\)
\(\Leftrightarrow\) cosα = \(\dfrac{1}{2}\)
* tanα = \(\dfrac{sin\alpha}{cos\alpha}=\dfrac{\sqrt{3}}{2}:\dfrac{1}{2}=\sqrt{3}\)
* tanα . cotα = 1
\(\Leftrightarrow\sqrt{3}.cot\alpha=1\)
\(\Leftrightarrow cot\alpha=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}\)
Tính \(tana+cota\) à bạn?