Đáp án: C+B
Giải thích các bước giải:
$\begin{array}{l}
20)\\
{\sin ^2}a + {\cos ^2}a = 1\\
{\sin ^2}a = {\cos ^2}\left( {{{90}^0} - a} \right)\\
\Leftrightarrow {\sin ^2}a + {\sin ^2}\left( {{{90}^0} - a} \right) = 1\\
A = {\sin ^2}{5^0} + {\sin ^2}{10^0} + ... + {\sin ^2}{90^0}\\
= \left( {{{\sin }^2}{5^0} + {{\sin }^2}{{85}^0}} \right) + \left( {{{\sin }^2}{{10}^0} + {{\sin }^2}{{80}^0}} \right)\\
+ ... + \left( {{{\sin }^2}{{40}^0} + {{\sin }^2}{{50}^0}} \right) + {\sin ^2}{45^0} + {\sin ^2}{90^0}\\
= 1 + 1 + ... + 1 + \dfrac{1}{2} + 1\\
= 1.8 + \dfrac{1}{2} + 1\\
= \dfrac{{19}}{2}\\
\Leftrightarrow C\\
21)\cot a = \tan \left( {{{90}^0} - a} \right)\\
\Leftrightarrow \cot a.\cot \left( {{{90}^0} - a} \right) = 1\\
T = \cot {5^0}.\cot {10^0}.\cot {15^0}...\cot {80^0}.\cot {85^0}\\
= \cot {5^0}.\cot {85^0}.\cot {10^0}.\cot {80^0}....\cot {45^0}\\
= 1.1...\cot {45^0}\\
= 1\\
\Leftrightarrow {T^2} = 1\\
\Leftrightarrow B
\end{array}$
