system of equation with cosh and sinh

Question

is there simple a way solve this system to find the unknown x and y

$$cosh\frac{a+x}y=\frac{b}{y}$$

$$sinh\frac{a+x}y=tanθ$$

My attemp:

dividing these equations we get

$$tanh\frac{a+x}y=y\frac{tanθ}{b}$$

Answer 1

Using the analytical expressions of $\cosh(x)$ and $\sinh(x)$.

First equation: $$ \exp(\frac{a+x}{y})+\exp(-\frac{a+x}{y}) = 2\frac{b}{y} $$ Second equation: $$ \exp(\frac{a+x}{y})-\exp(-\frac{a+x}{y}) = 2\tan(\theta) $$ Then, add the two equations and subtract the two equations to get new two equations: $$ \exp(\frac{a+x}{y}) = \frac{b}{y} + \tan(\theta) $$ and $$ \exp(-\frac{a+x}{y}) = \frac{b}{y} - \tan(\theta) $$ Multiply both equations, $$ 1 = \frac{b^2}{y^2} - \tan^2(\theta) $$ you will get $y = \pm b\cos(\theta)$, then substitute to get $x$

Answer 2

I forgot to say that the suggested answer is

$$y=bcosθ$$ and

$$x=-a +\frac{b}{2} cosθ *ln\frac{1+sinθ}{1-sinθ}$$