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I'm trying to prove that

$|\cot (x+iy)|^2=\frac{\cos^2x+\sinh^2y}{\sin^2x+\sinh^2y}$

for $x+iy \in \mathbb{C} \setminus \pi\mathbb{Z}$.

I've tried to use the identities $\cos(iy)=\cosh y$ and $\sin(iy)=i\sinh y$, but I obtain

$|\cot (x+iy)|^2=\frac{\cos^2(x) \cosh^2 y+\sin^2(x) \sinh^2y}{\sin^2(x)\cosh^2 y+\cos^2(x)\sinh^2y}$

What did I miss?

1 Answers 1

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You have missed nothing except simplification

Using $\cos^2x+\sin^2x=\cosh^2y-\sinh^2y=1$, eliminate $\cosh y,\sin x$ from the numerator and $\cosh y,\cos x$ from the denominator.

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

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

Again, $\cos^2x+\sinh^2y=1-\sin^2x+\cosh^2y-1=\cosh^2y-\sin^2x$

and $\sin^2x+\sinh^2y=1-\cos^2x+\cosh^2y-1=\cosh^2y-\cos^2x$(this is another simplified form)

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    I know, it was a typo .)2012-10-24