if sinh x=tan y, show that x=In(secy+tany)

Answer 1

sinh x = (e^x - e^-x)/2
So, now you have

e^x - e^-x = 2tan y
e^2x - 1 = e^x * 2tan y
e^2x - 2tany e^x - 1 = 0
Now use the quadratic formula to get

e^x = [2tany ±√(4tan^2y+4)]/2
= tany±√(tan^2y+1)
= tany±secy

x = ln(tany±secy)

Pick the principal branch of the curve.

Answer 2

if a=c cosh x and b=c sinh x, prove that(a+b)^2e^-2x=a^2-b^2

Answer 3

(a+b)^2e^-2x=a^2-b^2
divide both sides by a+b:
(a+b) e^-2x = a-b
(a+b)/(a-b) = e^2x

a+b = ce^x
a-b = ce^-x

and the rest follows.