Prove that 1+tan^2x/1+cot^2x=(1-tanx/1-cotx)^2

To prove : {tex} \frac{{1 + {{\tan }^2}\theta }}{{1 + {{\cot }^2}\theta }} = {\left( {\frac{{1 - \tan \theta }}{{1 - \cot \theta }}} \right)^2}{/tex}
Consider : {tex} \frac{{1 + {{\tan }^2}\theta }}{{1 + {{\cot }^2}\theta }} = \frac{{1 + \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}}{{1 + \frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}}}=\frac{{\frac{{{{\cos }^2}\theta + {{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}}{{\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\sin }^2}\theta }}}}{/tex}
{tex}= \frac{{\frac{1}{{{{\cos }^2}\theta }}}}{{\frac{1}{{{{\sin }^2}\theta }}}} = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}{/tex} {tex} \left[ {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right]{/tex}
{tex}= {\tan ^2}\theta {/tex}
Consider {tex} {\left( {\frac{{1 - \tan \theta }}{{1 - \cot \theta }}} \right)^2} = \frac{{1 + {{\tan }^2}\theta - 2\tan \theta }}{{1 + {{\cot }^2}\theta - 2\cot \theta }}{/tex}
{tex}= \frac{{{{\sec }^2}\theta - 2\tan \theta }}{{\cos e{c^2}\theta - 2\cot \theta }}{/tex} {tex} \left[ {\because 1 + {{\tan }^2}\theta = {{\sec }^2}\theta } \right]{/tex}
{tex}= \frac{{\frac{1}{{{{\cos }^2}\theta }} - \frac{{2\sin \theta }}{{\cos \theta }}}}{{\frac{1}{{{{\sin }^2}\theta }} - \frac{{2\cos \theta }}{{\sin \theta }}}} = \frac{{\frac{{1 - 2\sin \theta \cos \theta }}{{{{\cos }^2}\theta }}}}{{\frac{{1 - 2\sin \theta \cdot \cos \theta }}{{{{\sin }^2}\theta }}}}{/tex}
{tex} = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} = {\tan ^2}\theta {/tex}