Đáp án:
\[A = - \frac{2}{{15}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\
\Leftrightarrow {\left( {\frac{3}{5}} \right)^2} + {\cos ^2}\alpha = 1\\
\Leftrightarrow {\cos ^2}\alpha = {\left( {\frac{4}{5}} \right)^2}\\
\frac{\pi }{2} < \alpha < \pi \Rightarrow \cos \alpha < 0 \Rightarrow \cos \alpha = - \frac{4}{5}\\
\sin \alpha = cos\left( {\frac{\pi }{2} - \alpha } \right)\\
\sin \alpha = \sin \left( {\pi - \alpha } \right)\\
A = \cos \left( {\frac{\pi }{2} - \alpha } \right) + \sin \left( {2015\pi - \alpha } \right) + \cot \left( {2016\pi + \alpha } \right)\\
= \sin \alpha + \sin \left( {\pi - \alpha } \right) + \cot \alpha \\
= sin\alpha + sin\alpha + \frac{{\cos \alpha }}{{\sin \alpha }}\\
= \frac{3}{5}.2 - \frac{4}{3} = - \frac{2}{{15}}
\end{array}\)
