Prove trigonometry identity for $secxquad -sinx$

I'm trying to prove this equality but I' stuck at the second step.

Please give me some hints or other ways to proceed.

$\frac { { tan }^{ 2 }x\quad +\quad { cos }^{ 2 }x }{ sinx\quad +\quad secx } \quad \equiv \quad secx\quad -sinx\\ \\ sinx=x\\ cosx=y\\ \\ \frac { \frac { { x }^{ 2 } }{ { y }^{ 2 } } +\frac { { y }^{ 4 } }{ { y }^{ 2 } } }{ \frac { xy }{ y } +\frac { 1 }{ y } } \quad \equiv \quad \frac { 1 }{ y } -x\quad =\quad \frac { 1-xy }{ y } \quad \quad \quad \quad \quad (1)\\ \\ \\ \frac { \left( \frac { { x }^{ 2 }+{ y }^{ 4 } }{ { y }^{ 2 } } \right) }{ \left( \frac { xy+1 }{ y } \right) } \quad \equiv \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (2)\\ \\ \\ \frac { { x }^{ 2 }+{ y }^{ 4 } }{ y(xy+1) } \quad \equiv \qquad \qquad \qquad \qquad \qquad \qquad \qquad (3)$

Answer

The key is to see that $(x+y)(x-y) = x^2-y^2$.
\begin{align*}
\frac{\tan^2{x} + \cos^2{x}}{\sin x + \sec x} &= \frac{\tan^2{x} + 1 - 1 + \cos^2{x}}{\sin x + \sec x} \\
&= \frac{\sec^2{x} - \sin^2{x}}{\sin x + \sec x} \\
&= \sec x - \sin x.
\end{align*}