Need help with doing a hyperbolic trigonometry problem.

Question

If $x=2\cos(\alpha)\cosh(\beta)$ and $y=2\sin(\alpha)\sinh(\beta)$, prove: $$\sec(\alpha+i\beta)+\sec(\alpha-i\beta)=\frac{4x}{(x^2+y^2)}$$ I had an incorrect equation. The iota was missing.

Answer 1

$$\sec(\alpha+i\beta)+\sec(\alpha-i\beta)$$ $$\frac{1}{\cos(\alpha+i\beta)}+\frac{1}{\cos(\alpha-i\beta)}$$ $$\frac{2\cos(\alpha)\cos(i\beta)}{(\cos(\alpha)\cos(i\beta))^2-(\sin(\alpha)\sin(i\beta))^2}$$ $$\frac{\frac{x}{\cosh(\beta)}\cosh(\beta)}{(\frac{x}{2\cosh(\beta)}\cosh(\beta))^2-(\frac{y}{2\sinh(\beta)}i\sinh(\beta))^2}$$ $$\frac{4x}{x^2+y^2}$$