Simplify trigonometric expression of hyperbolic functions

I have $$\cos^2x\cosh^2y - \sin^2x\sinh^2y$$. I saw it written simplified as $$\cosh^2 y - \sin^2 x$$. But I don't get how to get it.

My attempts were to write $$\cosh^2y -1$$ instead of $$\sinh^2y$$ but that way I get $$\cos^2x\cosh^2y - \sin^2x\sinh^2y = \cosh^2y(\cos^2x-\sin^2x) + \sin^2x$$.

What am I doing wrong?

• Should it be $$\cos^2x\cosh^2y+\sin^2x\sinh^2y=(1-\sin^2x)\cosh^2y+\sin^2x(\cosh^2y-1)$$ – lab bhattacharjee Nov 12 '18 at 10:33
• then I have $\cos^2x\cosh^2y+\sin^2x\sinh^2y=(1-\sin^2x)\cosh^2y-\sin^2x(\cosh^2y-1) = \cosh^2y - \sin^2x\cosh^2 - \sin^2x\cosh^y + \sin^2x = \cosh^2y - 2\sin^2x\cosh^2 + \sin^2x$. Which is not equal to $\cosh^2y-\sin^2x$ – user3132457 Nov 12 '18 at 16:04
• seems like there should be + instead of -. In that case I'm getting the right answer. – user3132457 Nov 12 '18 at 16:22
• I was actually finding real, imaginary parts of $\tan(x+yi)$ and the modulus of it. I have $$Re(\tan(x+yi)) = {\frac{\sin2x}{\cos2x+\cosh2x}}, Im(\tan(x+yi))= {\frac{\sinh2y}{\cos2x+\cosh2x}}$$. But I can't seem to find the modulus. The answer is $${\sqrt{\frac{\cosh2y-\cos2x}{\cosh2y+\cosh2x}}}$$ – user3132457 Nov 12 '18 at 18:50
• Bottom of fraction should be $\cosh2y+\cos2x$ – user3132457 Nov 12 '18 at 19:14