# Question:solving sqrt(y)=tanh(x) for y

## Question:solving sqrt(y)=tanh(x) for y

Maple 2020

When solving this by hand   sqrt(y)=tanh(x), I would square both sides to obtain  y=tanh(x)^2 as solution.

Maple does this when RHS is tan,cos,sin, but not when RHS is hyperbolic.

Why is that, and how to best tell Maple to give  the simple solution? Here is an example

```restart;
eq:=sqrt(y)=tanh(x);
sol:=solve(eq,y);```

Compare to Mathematica:

I can post-process Maple solution like this

```restart;
eq:=sqrt(y)=tanh(x);
sol:=solve(eq,y);
sol:=convert(sol,trigh);
sol:=simplify(sol)```

Which is tanh(x)^2.

But this is all too much work, since this is done in a program, and I do not know what specific convert to use at the time to try.

My question is, why does not maple simply solves   sqrt(y) = f(x)  giving     y=f(x)^2?  it works for generic function

```restart;
eq:=sqrt(y)=f(x);
sol:=solve(eq,y);```

it works for non-hyper trig functions also

```restart;
eq:=sqrt(y)=tan(x);
sol:=solve(eq,y);```

Is there a trick to make it work for hyperbolic functions also automatically?

ps. for now, to avoid all these issues, I am doing this

```restart;
the_rhs:=tanh(x);
eq:=sqrt(y)=f;
sol:=solve(eq,y);
sol:=subs(f=the_rhs,sol);
```

Since I know that the RHS contains no y before hand. The above bypasses any issues Maple has with hyper trig functions.

﻿