\(\tan\alpha+\cot\alpha=2\) . Tính \(\alpha\)
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Lời giải:
a.
$\tan a+\cot a=2\Leftrightarrow \tan a+\frac{1}{\tan a}=2$
$\Leftrightarrow \frac{\tan ^2a+1}{\tan a}=2$
$\Leftrightarrow \tan ^2a-2\tan a+1=0$
$\Leftrightarrow (\tan a-1)^2=0\Rightarrow \tan a=1$
$\cot a=\frac{1}{\tan a}=1$
$1=\tan a=\frac{\cos a}{\sin a}\Rightarrow \cos a=\sin a$
Mà $\cos ^2a+\sin ^2a=1$
$\Rightarrow \cos a=\sin a=\pm \frac{1}{\sqrt{2}}$
b.
Vì $\sin a=\cos a=\pm \frac{1}{\sqrt{2}}$
$\Rightarrow \sin a\cos a=\frac{1}{2}$
$E=\frac{\sin a.\cos a}{\tan ^2a+\cot ^2a}=\frac{\frac{1}{2}}{1+1}=\frac{1}{4}$
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sin a=3/5
=>cos a=4/5
tan a=3/5:4/5=3/4; cot a=1:3/4=4/3
M=(4/3+3/4):(4/3-3/4)=25/7
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\(P=\dfrac{\dfrac{sina}{cosa}+\dfrac{cosa}{sina}}{\dfrac{sina}{cosa}-\dfrac{3cosa}{sina}}=\dfrac{sin^2a+cos^2a}{sin^2a-3cos^2a}=\dfrac{1}{sin^2a-3\left(1-sin^2a\right)}=\dfrac{1}{4sin^2a-3}=\dfrac{1}{4.\left(\dfrac{1}{3}\right)^2-3}=...\)
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Do \(\dfrac{\pi}{2}< \alpha< \pi\) nên \(tan\alpha< 0,cot\alpha< 0;cos\alpha< 0\).
Vì vậy: \(cos\alpha=-\sqrt{1-sin^2\alpha}=-\dfrac{\sqrt{7}}{4}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{3}{4}:\dfrac{-\sqrt{7}}{4}=\dfrac{-3}{\sqrt{7}}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{-\sqrt{7}}{3}\).
\(A=\dfrac{2tan\alpha-3cot\alpha}{cos\alpha+tan\alpha}\)\(=\dfrac{2.\dfrac{-3}{\sqrt{7}}-3.\dfrac{-\sqrt{7}}{3}}{\dfrac{-\sqrt{7}}{4}+\dfrac{-3}{\sqrt{7}}}\)
\(=\dfrac{\dfrac{-6}{\sqrt{7}}+\sqrt{7}}{\dfrac{-7-12}{4\sqrt{7}}}\)\(=\dfrac{\dfrac{-6+7}{\sqrt{7}}.4\sqrt{7}}{-19}\)\(=\dfrac{\dfrac{1}{\sqrt{7}}.4\sqrt{7}}{-19}=-\dfrac{4}{19}\).
b) \(\dfrac{cos^2\alpha+cot^2\alpha}{tan\alpha-cot\alpha}=\dfrac{\left(-\dfrac{\sqrt{7}}{4}\right)^2+\left(\dfrac{-\sqrt{7}}{3}\right)^2}{\dfrac{-3}{\sqrt{7}}+\dfrac{\sqrt{7}}{3}}\)
\(=\dfrac{\dfrac{7}{16}+\dfrac{7}{9}}{\dfrac{-9+7}{3\sqrt{7}}}=\dfrac{\dfrac{175}{144}}{\dfrac{-2}{3\sqrt{7}}}=\dfrac{-175}{96\sqrt{7}}\).
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a) \(tan^2\alpha+cot^2\alpha=\left(tan\alpha+cot\alpha\right)^2-2tan\alpha cot\alpha\)
\(=m^2-2\).
b) \(tan^3\alpha+cot^3\alpha=\left(tan\alpha+cot\alpha\right)\)\(\left(tan^2\alpha-tan\alpha cot\alpha+cot^2\alpha\right)\)
\(=m\left(tan^2\alpha+cot^2\alpha-tan\alpha cot\alpha\right)\)
\(=m\left(m^2-2-2\right)=m\left(m^2-3\right)\).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(0< a< 90^0\)
=>\(sina>0\)
\(sin^2a+cos^2a=1\)
=>\(sin^2a=1-\dfrac{9}{16}=\dfrac{7}{16}\)
=>\(sina=\dfrac{\sqrt{7}}{4}\)
\(tana=\dfrac{sina}{cosa}=\dfrac{\sqrt{7}}{4}:\dfrac{3}{4}=\dfrac{\sqrt{7}}{3}\)
\(cota=\dfrac{1}{tana}=\dfrac{3}{\sqrt{7}}\)
\(A=\dfrac{tana+3cota}{tana+cota}=\dfrac{\dfrac{\sqrt{7}}{3}+\dfrac{9}{\sqrt{7}}}{\dfrac{3}{\sqrt{7}}+\dfrac{\sqrt{7}}{3}}\)
\(=\dfrac{34}{3\sqrt{7}}:\dfrac{16}{3\sqrt{7}}=\dfrac{17}{8}\)
\(\tan\alpha+\cot\alpha=2\Leftrightarrow\tan\alpha+\frac{1}{\tan\alpha}=2\)
\(\Rightarrow\tan^2\alpha+1=2\tan\alpha\)
\(\Leftrightarrow\tan\alpha=1\Rightarrow\alpha=45\)