hyperbolic/trig question

[ggjwball]

if sinh x = tan y, show that tanh x = sin y.

Tried sinh (x) = (e^x - e^-x )/2 =tan(y) , so numerator is opposite side of a triangle and denominator is adjacent side, so it is possible to derive term for hypotenuse. Thus derive an expression for sin y. It then should be possible to match this expression to the exponential form of tanh(x)

 

[Deleted member 4993]

if sinh x = tan y, show that tanh x = sin y.

Tried sinh (x) = (e^x - e^-x )/2 =tan(y) , so numerator is opposite side of a triangle and denominator is adjacent side, so it is possible to derive term for hypotenuse. Thus derive an expression for sin y. It then should be possible to match this expression to the exponential form of tanh(x)

So then what did you get?

Where are you stuck?

 

[Ishuda]

if sinh x = tan y, show that tanh x = sin y.

Tried sinh (x) = (e^x - e^-x )/2 =tan(y) , so numerator is opposite side of a triangle and denominator is adjacent side, so it is possible to derive term for hypotenuse. Thus derive an expression for sin y. It then should be possible to match this expression to the exponential form of tanh(x)

Use the identities involving sinh²(x) and cosh²(x) and the identities involving tan²(y) and sec²(y) = 1/cos²(y)