Đáp án:
$\begin{array}{l}
a)\sin \alpha + \cos \alpha = \sqrt 2 \\
\Rightarrow \sqrt 2 .\sin \left( {\alpha + \dfrac{\pi }{4}} \right) = \sqrt 2 \\
\Rightarrow \sin \left( {\alpha + \dfrac{\pi }{4}} \right) = 1\\
\Rightarrow \alpha + \dfrac{\pi }{4} = \dfrac{\pi }{2} + k2\pi \\
\Rightarrow \alpha = \dfrac{\pi }{4} + k2\pi \\
\Rightarrow \alpha = \dfrac{\pi }{4}\left( {do:0 < \alpha < \dfrac{\pi }{2}} \right)\\
b)\sin \alpha + \cos \alpha = \dfrac{{\sqrt 3 }}{4}\\
\Rightarrow \sqrt 2 .\sin \left( {\alpha + \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 3 }}{4}\\
\Rightarrow \sin \left( {\alpha + \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 6 }}{8}\\
\Rightarrow \left[ \begin{array}{l}
\alpha + {45^0} = {18^0} + k{.360^0}\\
\alpha + {45^0} = {180^0} - {18^0} + k{.360^0}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\alpha = - {27^0} + k{.360^0}\\
\alpha = {117^0} + k{.360^0}
\end{array} \right.\\
\Rightarrow \alpha \in \emptyset \left( {do:0 < \alpha < {{90}^0}} \right)\\
c)Dkxd:\left\{ \begin{array}{l}
\sin \alpha \ne 0\\
\cos \alpha \ne 0
\end{array} \right. \Rightarrow a \ne \dfrac{{k\pi }}{2}\\
\tan \alpha + \cot \alpha = 2\\
\Rightarrow \tan \alpha + \dfrac{1}{{\tan \alpha }} = 2\\
\Rightarrow {\tan ^2}\alpha - 2\tan \alpha + 1 = 0\\
\Rightarrow \tan \alpha = 1\\
\Rightarrow \alpha = \dfrac{\pi }{4} + k\pi \\
\Rightarrow \alpha = \dfrac{\pi }{4}
\end{array}$
