If xsinh w + ycosh w =1 and xcosh w - ysinh w=1, how do you show that (x^2 + y^2)^2 - 4xy = 0?

Aug 16, 2016

Refer to the Proof furnished in Explanation below.

Explanation:

We solve the given eqns. for $\cosh w$, &, $\sinh w$ and, then, we

will use the Identity ${\cosh}^{2} w - {\sinh}^{2} w = 1. \ldots \ldots \ldots . . \left(\star\right)$.

For the sake of brevity, we write $\cosh w = c$ and $\sinh w = s$ in the

given eqns. and, solve them for $c \mathmr{and} s$ by Kramer's Method.

The eqns. are $c y + s x - 1 = 0 , \mathmr{and} , c x - s y - 1 = 0$.

$\therefore \frac{c}{| \left(x , - 1\right) , \left(- y , - 1\right) |} = - \frac{s}{| \left(y , - 1\right) , \left(x , - 1\right) |} = \frac{1}{| \left(y , x\right) , \left(x , - y\right) |}$.

$\therefore \frac{c}{-} \left(x + y\right) = - \frac{s}{x - y} = \frac{1}{-} \left({x}^{2} + {y}^{2}\right)$.

$\therefore c = \cosh w = \frac{x + y}{{x}^{2} + {y}^{2}} , s = \sinh w = \frac{x - y}{{x}^{2} + {y}^{2}}$.

By $\left(\star\right)$, then, we have,

$\frac{{\left(x + y\right)}^{2} - {\left(x - y\right)}^{2}}{{x}^{2} + {y}^{2}} ^ 2 = 1 , \mathmr{and} , \frac{4 x y}{{x}^{2} + {y}^{2}} ^ 2 = 1$, i.e.,

${\left({x}^{2} + {y}^{2}\right)}^{2} - 4 x y = 0$, as desired!.

Hence, the Proof.

Enjoy Maths.!