Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
\dfrac{\pi }{2} < \alpha < \pi \Rightarrow \left\{ \begin{array}{l}
\sin \alpha > 0\\
cos\alpha < 0
\end{array} \right.\\
{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\
\Leftrightarrow {\left( {\dfrac{1}{2}} \right)^2} + {\cos ^2}\alpha = 1\\
\Leftrightarrow {\cos ^2}\alpha = \dfrac{3}{4}\\
\cos \alpha < 0 \Rightarrow \cos \alpha = - \dfrac{{\sqrt 3 }}{2}\\
\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }} = - \dfrac{1}{{\sqrt 3 }}\\
\cot \alpha = \dfrac{{\cos \alpha }}{{\sin \alpha }} = - \sqrt 3 \\
2,\\
\cos \left( {\dfrac{\pi }{2} - a} \right).\sin \left( {\dfrac{\pi }{2} - b} \right) - \sin \left( {a - b} \right)\\
= \sin a.\cos b - \left( {\sin a.\cos b - \cos a.\sin b} \right)\\
= \sin a.\cos b - \sin a.\cos b + \cos a.\sin b\\
= \cos a.\sin b\\
3,\\
\cos \left( {a + b} \right).\cos \left( {a - b} \right)\\
= \dfrac{1}{2}.\left[ {\cos \left( {a + b + a - b} \right) + \cos \left( {a + b - a + b} \right)} \right]\\
= \dfrac{1}{2}\left[ {\cos 2a + \cos 2b} \right]\\
= \dfrac{1}{2}.\left[ {\left( {2{{\cos }^2}a - 1} \right) - \left( {1 - 2{{\sin }^2}b} \right)} \right]\\
= \dfrac{1}{2}.\left( {2{{\cos }^2}a + 2{{\sin }^2}b} \right)\\
= {\cos ^2}a + {\sin ^2}b
\end{array}\)
