Đáp án:
\[P = \dfrac{{29}}{5}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{\pi }{2} < a < \pi \Rightarrow \left\{ \begin{array}{l}
\sin a > 0\\
\cos a < 0
\end{array} \right.\\
{\sin ^2}a + {\cos ^2}a = 1\\
\cos a < 0 \Rightarrow \cos a = - \sqrt {1 - {{\sin }^2}a} = - \sqrt {1 - {{\left( {\dfrac{3}{5}} \right)}^2}} = - \dfrac{4}{5}\\
P = 1 - 2{\sin ^2}\left( {\dfrac{\pi }{4} - a} \right) + \sin 2a + \cos \left( {\pi - 2a} \right) - 6\tan \left( {\dfrac{\pi }{2} - a} \right)\\
= \left[ {1 - 2{{\sin }^2}\left( {\dfrac{\pi }{4} - a} \right)} \right] + 2\sin a.\cos a - \cos 2a - 6\cot a\\
= \cos \left( {\dfrac{\pi }{2} - 2a} \right) + 2\sin a.\cos a - \left( {2{{\cos }^2}a - 1} \right) - \dfrac{{6\cos a}}{{\sin a}}\\
= \sin 2a + 2\sin a.\cos a - 2{\cos ^2}a + 1 - \dfrac{{6\cos a}}{{\sin a}}\\
= 4\sin a.\cos a - 2{\cos ^2}a - \dfrac{{6\cos a}}{{\sin a}} + 1\\
= 4.\dfrac{3}{5}.\left( { - \dfrac{4}{5}} \right) - 2.{\left( { - \dfrac{4}{5}} \right)^2} - \dfrac{{6.\left( { - \dfrac{4}{5}} \right)}}{{\dfrac{3}{5}}} + 1\\
= \dfrac{{29}}{5}
\end{array}\)
