Đáp án:
$\begin{array}{l}
a)f\left( x \right) = \frac{{\sin x + 1}}{{{\mathop{\rm s}\nolimits} {\rm{inx}} - 1}}\\
ĐKxđ:{\mathop{\rm s}\nolimits} {\rm{inx}} - 1 \ne 0\\
\Rightarrow {\mathop{\rm s}\nolimits} {\rm{inx}} \ne 1\\
\Rightarrow x \ne \frac{\pi }{2} + k2\pi \left( {k \in Z} \right)\\
b)f\left( x \right) = \frac{{2\tan x + 2}}{{{\rm{cosx - 1}}}}\\
Đkxđ{\rm{d}}:\left\{ \begin{array}{l}
{\rm{cosx}} \ne {\rm{0}}\\
{\rm{cosx}} \ne {\rm{1}}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x \ne \frac{\pi }{2} + k\pi \\
x \ne k2\pi
\end{array} \right.\\
c)f\left( x \right) = \frac{{\cot x}}{{{\mathop{\rm s}\nolimits} {\rm{inx}} + 1}}\\
Đkxđ:\left\{ \begin{array}{l}
{\mathop{\rm s}\nolimits} {\rm{inx}} \ne 0\\
{\mathop{\rm s}\nolimits} {\rm{inx}} \ne - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x \ne k\pi \\
x \ne \frac{{ - \pi }}{2} + k2\pi
\end{array} \right.
\end{array}$
