biết cot\(\alpha\)=\(\sqrt{5}\). tính A=\(\frac{\sin\alpha^2+\cos\alpha^2}{\sin\alpha.\cos\alpha}\)
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ND
Nguyễn Đức Trí
VIP
12 tháng 9 2023
1) \(cot\alpha=\sqrt[]{5}\Rightarrow tan\alpha=\dfrac{1}{\sqrt[]{5}}\)
\(C=sin^2\alpha-sin\alpha.cos\alpha+cos^2\alpha\)
\(\Leftrightarrow C=\dfrac{1}{cos^2\alpha}\left(tan^2\alpha-tan\alpha+1\right)\)
\(\Leftrightarrow C=\left(1+tan^2\alpha\right)\left(tan^2\alpha-tan\alpha+1\right)\)
\(\Leftrightarrow C=\left(1+\dfrac{1}{5}\right)\left(\dfrac{1}{5}-\dfrac{1}{\sqrt[]{5}}+1\right)\)
\(\Leftrightarrow C=\dfrac{6}{5}\left(\dfrac{6}{5}-\dfrac{\sqrt[]{5}}{5}\right)=\dfrac{6}{25}\left(6-\sqrt[]{5}\right)\)
12 tháng 9 2023
1: \(cota=\sqrt{5}\)
=>\(cosa=\sqrt{5}\cdot sina\)
\(1+cot^2a=\dfrac{1}{sin^2a}\)
=>\(\dfrac{1}{sin^2a}=1+5=6\)
=>\(sin^2a=\dfrac{1}{6}\)
\(C=sin^2a-sina\cdot\sqrt{5}\cdot sina+\left(\sqrt{5}\cdot sina\right)^2\)
\(=sin^2a\left(1-\sqrt{5}+5\right)=\dfrac{1}{6}\cdot\left(6-\sqrt{5}\right)\)
2: tan a=3
=>sin a=3*cosa
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>\(\dfrac{1}{cos^2a}=1+9=10\)
=>\(cos^2a=\dfrac{1}{10}\)
\(B=\dfrac{3\cdot cosa-cosa}{27\cdot cos^3a+3\cdot cos^3a+2\cdot3\cdot cosa}\)
\(=\dfrac{2\cdot cosa}{30cos^3a+6cosa}=\dfrac{2}{30cos^2a+6}\)
\(=\dfrac{2}{3+6}=\dfrac{2}{9}\)
AT
29 tháng 7 2021
Ta có: \(cot\alpha=\dfrac{cos\alpha}{sin\alpha}=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}\)
Lại có: \(\dfrac{1}{cot\alpha}=tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{sin^2\alpha}{cos\alpha.sin\alpha}=\dfrac{1}{\sqrt{5}}\)
\(\Rightarrow A=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}+\dfrac{sin^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}+\dfrac{1}{\sqrt{5}}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)
29 tháng 7 2021
Ta có : cot α = \(\sqrt{5}\Rightarrow\dfrac{cos\alpha}{sin\alpha}=\sqrt{5}\Rightarrow cos\alpha=\sqrt{5}.sin\alpha\)
\(A=\dfrac{sin^2\alpha+cos^2\alpha}{sin\alpha.cos\alpha}\)
\(A=\dfrac{sin^2\alpha+\left(\sqrt{5}sin\alpha\right)^2}{sin\alpha.\sqrt{5}sin\alpha}=\dfrac{sin^2\alpha+5sin^2\alpha}{\sqrt{5}sin^2\alpha}\)
\(A=\dfrac{6sin^2\alpha}{\sqrt{5}sin^2\alpha}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 3 2022
\(tana-5cota+4=0\Rightarrow tana-\dfrac{5}{tana}+4=0\)
\(\Rightarrow tan^2a+4tana-5=0\Rightarrow\left[{}\begin{matrix}tana=1\\tana=-5\end{matrix}\right.\)
\(A=\dfrac{4sina+2cosa}{3sina-cosa}=\dfrac{\dfrac{4sina}{cosa}+\dfrac{2cosa}{cosa}}{\dfrac{3sina}{cosa}-\dfrac{cosa}{cosa}}=\dfrac{4tana+2}{3tana-1}=\left[{}\begin{matrix}3\\\dfrac{9}{8}\end{matrix}\right.\)
27 tháng 8 2021
a/ \(A=\frac{cot^2a-cos^2a}{cot^2a}-\frac{sina.cosa}{cota}\)
\(=\frac{\frac{cos^2a}{sin^2a}-cos^2a}{\frac{cos^2a}{sin^2a}}-\frac{sina.cosa}{\frac{cosa}{sina}}\)
\(=\left(1-sin^2a\right)-sin^2a=1\)
27 tháng 8 2021
b/ \(B=\left(cosa-sina\right)^2+\left(cosa+sina\right)^2+cos^4a-sin^4a-2cos^2a\)
\(=cos^2a-2cosa.sina+sin^2a+cos^2a+2cosa.sina+sin^2a+\left(cos^2a+sin^2a\right)\left(cos^2a-sin^2a\right)-2cos^2a\)
\(=2+\left(cos^2a-sin^2a\right)-2cos^2a\)
\(=2-sin^2a-cos^2a=2-1=1\)
\(\cot\alpha=\frac{\cos\alpha}{\sin\alpha}=\sqrt{5}\Rightarrow\frac{\cos\alpha}{\sqrt{5}}=\frac{\sin\alpha}{1}\)
Đặt \(\frac{\cos\alpha}{\sqrt{5}}=\frac{\sin\alpha}{1}=k\)thì \(\cos\alpha=\sqrt{5}k,\sin\alpha=k\)
Vậy \(A=\frac{\sin^2a+\cos^2\alpha}{\sin\alpha.\cos\alpha}=\frac{k^2+5k^2}{\sqrt{5}k.k}=\frac{6}{\sqrt{5}}\)