# Proof that $\cos^2(x)\cosh^2(y) + \sin^2(x)\sinh^2(y) = -1 + \sin^2(x) - \sinh^2(y)$

Could anyone offer a proof that

$$\cos^2(x)\cosh^2(y) + \sin^2(x)\sinh^2(y) = -1 + \sin^2(x) - \sinh^2(y)?$$

• Perhaps someone can offer a proof, but not a correct proof, because it isn't true. Consider the case $x=y=0$. – David Aug 4 '14 at 12:22

Note that $\sin^2 x+\cos^2 x=1$ and $\sinh^2 x=1+\cosh^2 x$
$$\cos^2 x \cosh^2 x+ \sin^2 x \sinh^2 x= (1-\sin^2 x) (1+\sinh^2 x)+ \sin^2 x \sinh^2 x= 1-\sin^2 x +\sinh^2 x$$