Đáp án:
$\begin{array}{l}
a){90^0} < x < {180^0}\\
\Rightarrow {\mathop{\rm cosx}\nolimits} < 0\\
\Rightarrow cosx = - \sqrt {1 - {{\sin }^2}x} = - \sqrt {1 - \dfrac{1}{4}} = - \dfrac{{\sqrt 3 }}{2}\\
\Rightarrow \tan x = \dfrac{{\sin x}}{{\cos x}} = - \dfrac{{\sqrt 3 }}{3}\\
\Rightarrow \cot x = - \sqrt 3 \\
b){180^0} < x < {270^0}\\
\Rightarrow \sin x < 0\\
\Rightarrow \sin x = - \sqrt {1 - {{\cos }^2}x} = - \dfrac{{12}}{{13}}\\
\Rightarrow \left\{ \begin{array}{l}
\tan x = \dfrac{{12}}{5}\\
{\mathop{\rm cost}\nolimits} = \dfrac{5}{{12}}
\end{array} \right.\\
c)\pi < x < \dfrac{{3\pi }}{2}\\
\Rightarrow \cos x < 0\\
Do:\dfrac{1}{{{{\cos }^2}x}} = {\tan ^2}x + 1 = 10\\
\Rightarrow \cos x = - \dfrac{{\sqrt {10} }}{{10}}\\
\Rightarrow \sin x = \cos x.tanx = - \dfrac{{3\sqrt {10} }}{{10}}\\
\Rightarrow \cot x = \dfrac{1}{3}\\
d)\dfrac{\pi }{2} < x < \pi \\
\Rightarrow \sin x > 0\\
\dfrac{1}{{{{\sin }^2}x}} = \cot {x^2} + 1 = 4\\
\Rightarrow \sin x = \dfrac{1}{2}\\
\Rightarrow \cos x = \sin x.\cot x = - \dfrac{{\sqrt 3 }}{2}\\
\Rightarrow \tan x = - \dfrac{{\sqrt 3 }}{3}
\end{array}$
