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