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Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 4 / 11E / b / Solution

The hyperbolic law of sines is

\frac{s i n h a}{s i n α} = \frac{s i n h b}{s i n β} = \frac{s i n h c}{s i n γ} .

Set

γ = π /2

in the cosine laws for sides and angles and eliminate

cos α

and

cos β

. Straight algebra gives the right-triangle identities

tan α = \frac{sinh a}{cosh a sinh b}, tan α tan β cosh c = 1 .

Solved by gpt-5.6-sol high.

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