$\dfrac{1+tan^4x}{tan^2x+cot^2x}=tan^2x$

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