bt tan alpha + cot alpha=2 tinh sin alpha , cos alpha 2sin^2 alpha – 3sin alpha -2 =0 tinh tan alpha bai 6 tinh cac ti so luong giac goc 15

Đáp án:

$\begin{array}{l}
1)\\
{\mathop{\rm tana}\nolimits}  + cota = 2\\
 \Rightarrow \tan a + \dfrac{1}{{{\mathop{\rm tana}\nolimits} }} = 2\\
 \Rightarrow {\tan ^2}a – 2{\mathop{\rm tana}\nolimits}  + 1 = 0\\
 \Rightarrow {\left( {\tan a – 1} \right)^2} = 0\\
 \Rightarrow {\mathop{\rm tana}\nolimits}  = 1\\
 \Rightarrow \dfrac{{\sin a}}{{\cos a}} = 1 \Rightarrow \sin a = \cos a\\
Do:\dfrac{1}{{{{\cos }^2}a}} = {\tan ^2}a + 1 = 2\\
 \Rightarrow {\cos ^2}a = \dfrac{1}{2}\\
 \Rightarrow \left[ \begin{array}{l}
\cos a = \sin a = \dfrac{{\sqrt 2 }}{2}\\
\cos a = \sin a =  – \dfrac{{\sqrt 2 }}{2}
\end{array} \right.\\
b)2{\sin ^2}a – 3\sin a – 2 = 0\\
 \Rightarrow \left( {2\sin a + 1} \right)\left( {\sin a – 2} \right) = 0\\
 \Rightarrow \sin a =  – \dfrac{1}{2}\\
 \Rightarrow {\cos ^2}a = 1 – {\sin ^2}a = \dfrac{3}{4}\\
 \Rightarrow {\tan ^2}a = \dfrac{1}{{{{\cos }^2}a}} – 1 = \dfrac{1}{3}\\
 \Rightarrow \left[ \begin{array}{l}
\tan a = \dfrac{{\sqrt 3 }}{3}\\
\tan a =  – \dfrac{{\sqrt 3 }}{3}
\end{array} \right.\\
B6)\\
 + \sin {15^0} = \dfrac{{\sqrt 6  – \sqrt 2 }}{4}\\
 + \cos {15^0} = \dfrac{{\sqrt 6  + \sqrt 2 }}{4}\\
 + \tan {15^0} = 2 – \sqrt 3 \\
 + \cot {15^0} = 2 + \sqrt 3 
\end{array}$