I'm trying to prove this equality but I' stuck at the second step.
Please give me some hints or other ways to proceed.
\begin{gather}\frac{\tan^2x + \cos^2x}{\sin x+ \sec x} \equiv \sec x - \sin x \\ \sin x = 0 \\ \cos x = y \\ \frac{\frac{x^2}{y^2}+ \frac{y^4}{y^2}}{\frac{xy}{y} + \frac{1}{y}} \equiv \frac{1}{y} - x = \frac{1-xy}{y} \tag{1} \\ \frac{ \frac{x^2+y^4}{y^2} }{ \frac{xy+1}{y} }\equiv \tag{2} \\ \frac{x^2+y^4}{y(xy+1)} \equiv\tag{3} \end{gather}
Start by multiplying both sides by $\sin x + \sec x$ to get:
$$\tan^2 x + \cos^2 x = (\sec x - \sin x)(\sec x + \sin x)$$
And work the rest out from there :)
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*}
$$\cos^2(x)-1=-\sin^2(x)$$ $$\tan^2(x)=\sec^2(x)-1$$
$$\frac{\tan^2(x)+\cos^2(x)}{\sin(x)+\sec(x)}=\frac{\left(\sec^2(x)-1 \right)+\cos^2(x)}{\sin(x)+\sec(x)}$$
$$=\frac{\sec^2(x)-\sin^2(x)}{\sin(x)+\sec(x)}=\frac{(\sin(x)+\sec(x))(\sin(x)-\sec(x))}{\sin(x)+\sec(x)}$$
I think you can do the rest from here. Though try not to substitute a function for the same variable (e.g. use $t=\cos(x)$ not $x=\cos(x)$)
Suggestion: Try writing the right hand side as a fraction over the same denominator, $\sin(x)+\sec(x)$. Then see if you can manipulate the numerators to be the same.
