Is there an easy way to simplify $\tanh(2\operatorname{arctanh}(x))$ and the like?

Question

Is there an easy way to generally simplify any hyperbolic functions of inverse hyperbolic functions, with examples shown below?

$$\tanh(2\operatorname{arctanh}(x))$$

$$\coth(\operatorname{arccosech}(x))$$

I'm wondering this as for my calculas exam in a week sometimes I am given questions of this format and we have not formally been taught how to do something like this.

I'm aware that $\cosh(\operatorname{arccosh}(x))=x$, etc., but I find that I can't do examples similar to those above without wasting alot of my exam time.

Sorry about the wording of this question, I'm finding it very awkward to phrase.

Thanks in advance.

Answer 1

For the first one, use the double angle formula $$\tanh2A=\frac{2\tanh A}{1+\tanh^2A}$$ So the result is $$\frac {2x}{1+x^2}$$

For the second one, and similar expressions, you could try using the hyperbolic analogue of the trig mnemonic SOHCAHTOA which is CHASOATOH:

You have $$\operatorname{cosech}A=x$$ which means the adjacent side of the right angled triangle is $x$ and the opposite side is $1$ (following SOA). The hypotenuse is obviously $\sqrt{1+x^2}$ and therefore $\coth A$ is H/O which gives the result $\sqrt{1+x^2}$