$x$ in terms of $\tan(y)$ ?

$\begin{aligned} \sec y + \tan y & = x \\ \sqrt{1+\tan^2 y} + \tan y & = x \\ \sqrt{1+\tan^2 y} & = x - \tan y & \small \color{#3D99F6} \text{Squaring both sides.} \\ 1 + \cancel{\tan^2 y} & = x^2 - 2x \tan y + \cancel{\tan^2 y} \\ 2x \tan y & = x^2 - 1 \\ \implies \tan y & = \frac x2 - \frac 1{2x} \end{aligned}$

Therefore, $A-C+B =2-2+1 = \boxed 1$ .

The identity $\sec^2(y)-\tan^2(y)=1$ is particularly useful in this instance. Given that $x=\sec(y)+\tan(y)$ , we can use difference of the squares to express $\sec^2(y)-\tan^2(y)=1$ as $( \sec(y)+\tan(y) ) ( \sec(y)-\tan(y) ) = 1$ , which is $x (\sec(y)-\tan(y)) = 1$ . Rearranging $\sec(y)+\tan(y) = x$ , and substituting $x-\tan(y)$ for $\sec(y)$ in $x (\sec(y)-\tan(y)) = 1$ , we get $x (x-2\tan(y)) = 1$

Simplifying this, we get $\tan(y)$ = $\frac{x}{2}$ - $\frac{1}{2x}$ , where A is 2, C is 2, and B is 1, therefore A-C+B= 1