Đáp án:
$\begin{array}{l}
Do:\left\{ \begin{array}{l}
{\sin ^2}a + {\cos ^2}a = 1\\
\sin a = \cos \left( {{{90}^0} - a} \right)
\end{array} \right.\\
A = {\sin ^2}{17^0} + {\sin ^2}{24^0} + {\sin ^2}{75^0} + {\sin ^2}{66^0}\\
= {\sin ^2}{17^0} + {\sin ^2}{75^0} + \left( {{{\sin }^2}{{24}^0} + {{\sin }^2}{{66}^0}} \right)\\
= {\sin ^2}{17^0} + {\sin ^2}{75^0} + \left( {{{\sin }^2}{{24}^0} + {{\cos }^2}{{24}^0}} \right)\\
= 1,018 + 1\\
= 2,018\\
B = \sin {42^0} + \tan {45^0} - \cos {48^0}\\
= \sin {42^0} - \sin {42^0} + \tan {45^0}\\
= \tan {45^0}\\
= 1\\
C = \cot {27^0}.\cot {60^0}.\cot {63^0} + {\sin ^2}{44^0} + {\sin ^2}{46^0}\\
= \cot {60^0}.\tan {63^0}.\cot {63^0} + {\sin ^2}{44^0} + {\cos ^2}{44^0}\\
= \dfrac{1}{{\sqrt 3 }} + 1\\
= \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 }}\\
= \dfrac{{3 + \sqrt 3 }}{3}\\
D = \dfrac{{\sin {{48}^0}}}{{\cos {{42}^0}}} + \cot {60^0} + \tan {27^0}.\tan {63^0}\\
= 1 + \dfrac{1}{{\sqrt 3 }} + 1\\
= 2 + \dfrac{1}{{\sqrt 3 }}\\
= \dfrac{{2\sqrt 3 + 1}}{{\sqrt 3 }}\\
= \dfrac{{6 + \sqrt 3 }}{3}
\end{array}$
