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$\displaystyle \frac{ \cot x}{1 - \sin ^2 x} { ? = } \frac{ \tan x}{1 - \cos ^2 x}$
$\displaystyle \frac{ \cot x}{ \cos ^2 x} { ? = } \frac{ \tan x}{\sin ^2 x}$
$\displaystyle \frac{1}{\tan x} \frac{ 1}{ \cos ^2 x} { ? = } \frac{ \tan x}{\sin ^2 x}$
where $\displaystyle ? = $ is a question - is this true. Do you see it from here?