Đáp án:
$\begin{array}{l}
+ ){\left( {\frac{{\sin x + \frac{{\cos x}}{{\sin x}}}}{{1 + \sin x.\frac{{\sin x}}{{\cos x}}}}} \right)^2} = \frac{{{{\left( {{{\sin }^2}x + \cos x} \right)}^2}}}{{{{\sin }^2}x}}:\frac{{{{\left( {\cos x + {{\sin }^2}x} \right)}^2}}}{{{{\cos }^2}x}}\\
= \frac{{{{\left( {{{\sin }^2}x + \cos x} \right)}^2}}}{{{{\sin }^2}x}}.\frac{{{{\cos }^2}x}}{{{{\left( {{{\sin }^2}x + \cos x} \right)}^2}}} = \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}}\\
+ )\frac{{{{\sin }^2}x + {{\cot }^2}x}}{{1 + {{\sin }^2}.{{\tan }^2}x}} = \frac{{{{\sin }^2}x + \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}}}}{{1 + {{\sin }^2}x.\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}} = \frac{{\frac{{{{\sin }^4}x + {{\cos }^2}x}}{{{{\sin }^2}x}}}}{{\frac{{{{\cos }^2}x + {{\sin }^4}x}}{{{{\cos }^2}x}}}} = \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}}\\
\Rightarrow \frac{{{{\left( {\sin x + \cot x} \right)}^2}}}{{{{\left( {1 + \sin x.\tan x} \right)}^2}}} = \frac{{{{\sin }^2}x + {{\cot }^2}x}}{{1 + {{\sin }^2}.{{\tan }^2}x}}\left( { = \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}}} \right)
\end{array}$
