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

Question

1139 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-16T12:42:38

Refer to the Proof furnished in Explanation below.

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

will use the Identity #cosh^2w-sinh^2w=1............(star)#.

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

given eqns. and, solve them for #c and s# by Kramer's Method.

The eqns. are #cy+sx-1=0, and, cx-sy-1=0#.

#:. c/(|(x,-1),(-y,-1)|)=-s/(|(y,-1),(x,-1)|)=1/(|(y,x),(x,-y)|)#.

#:. c/-(x+y)=-s/(x-y)=1/-(x^2+y^2)#.

#:. c=cosh w=(x+y)/(x^2+y^2), s=sinh w=(x-y)/(x^2+y^2)#.

By #(star)#, then, we have,

#{(x+y)^2-(x-y)^2}/(x^2+y^2)^2=1, or, (4xy)/(x^2+y^2)^2=1#, i.e.,

#(x^2+y^2)^2-4xy=0#, as desired!.

Hence, the Proof.

Enjoy Maths.!