Looked for similar examples in IntMath, but couldn't find any.

X

can you help with following equation
simplify
1+sinh2A+cosh2A / 1-sinh2A-cosh2A

Relevant page
<a href="/analytic-trigonometry/5-trigonometric-equations.php">5. Trigonometric Equations</a>
What I've done so far
Looked for similar examples in IntMath, but couldn't find any.

Your expression is actually related to Exponential Functions rather than trigonometric ones. I don't include the topic of Hyperbolic Functions on my site yet, but you may find the following helpful:

Simplify this, then do a similar thing for the bottom of the fraction.

Then see what you are left with, after simplifying the whole thing.

I hope that makes sense.

Regards

X

Hello Lee
Your expression is actually related to Exponential Functions rather than trigonometric ones. I don't include the topic of Hyperbolic Functions on my site yet, but you may find the following helpful:
<a href="https://en.wikipedia.org/wiki/Hyperbolic_trigonometric_function">Hyperbolic function (Wikipedia)</a>
BTW, you are missing brackets in your question. The brackets make a huge difference to the meaning of the algebra. I think you meant this:
(1+sinh2A+cosh2A) / (1-sinh2A-cosh2A)
Using the math input system, it would look like this:
`(1+sinh2A+cosh2A) / (1-sinh2A-cosh2A)`
Without brackets, you get this:
`1+sinh2A+cosh2A / 1-sinh2A-cosh2A `
Anyway, to get you started, you need to know that
`sinh x = (e^x-e^-x)/2`
and
`cosh x = (e^x+e^-x)/2`
So the top line of your fraction will be:
`1+sinh2A+cosh2A = 1 + (e^(2A)-e^(-2A))/2 + (e^(2A)+e^(-2A))/2`
Simplify this, then do a similar thing for the bottom of the fraction.
Then see what you are left with, after simplifying the whole thing.
I hope that makes sense.
Regards