Đáp án:
$\begin{array}{l}
1)\pi < a < \dfrac{{3\pi }}{2}\\
\Leftrightarrow \cos a < 0\\
Do:\dfrac{1}{{{{\cos }^2}a}} = {\tan ^2}a + 1\\
\Leftrightarrow \dfrac{1}{{{{\cos }^2}a}} = {3^2} + 1 = 10\\
\Leftrightarrow {\cos ^2}a = \dfrac{1}{{10}}\\
\Leftrightarrow \cos a = - \dfrac{{\sqrt {10} }}{{10}}\\
Vậy\,\cos a = \dfrac{{ - \sqrt {10} }}{{10}}\\
2)\sin \left( {\dfrac{\pi }{2} - a} \right) + 2\cos \left( { - a} \right) - 4\cos \left( {a + \pi } \right)\\
= \cos a + 2\cos a + 4\cos a\\
= 7\cos a\\
3)\\
{\left( {\dfrac{{\sin a + \tan a}}{{\cos a + 1}}} \right)^2}\\
= {\left( {\dfrac{{\sin a + \dfrac{{\sin a}}{{\cos a}}}}{{\cos a + 1}}} \right)^2}\\
= {\left( {\sin a.\dfrac{{\dfrac{{\cos a + 1}}{{\cos a}}}}{{\cos a + 1}}} \right)^2}\\
= {\left( {\sin a.\dfrac{1}{{\cos a}}} \right)^2}\\
= {\tan ^2}a\\
4)\sin x = \dfrac{3}{5}\\
\dfrac{\pi }{2} < x < \pi \Leftrightarrow \cos x < 0\\
\Leftrightarrow \cot x < 0\\
Do:\dfrac{1}{{{{\sin }^2}a}} = {\cot ^2}a + 1\\
\Leftrightarrow {\cot ^2}a = \dfrac{{25}}{9} - 1 = \dfrac{{16}}{9}\\
\Leftrightarrow \cot a = - \dfrac{4}{3}\\
Vậy\,\cot a = \dfrac{{ - 4}}{3}
\end{array}$
