```
sage: u,r,t,a = var('u','r','t','a')
sage: g = (sqrt(a^2 + (r+t)^2) - sqrt(a^2 + (r-t)^2))/(2*r)
sage: E = u == g ; E
u == 1/2*(sqrt(a^2 + (r + t)^2) - sqrt(a^2 + (r - t)^2))/r
```

This can't be solved "automagically" by Sage's default solver (i. e. Maxima's).

However, there are some worarounds. First, one can "pilot" Sage's default solver to the solution. First, we try to eliminate the radicals by squaring :

```
sage: E2 = (E^2).expand(); E2
u^2 == 1/2*a^2/r^2 + 1/2*t^2/r^2 - 1/2*sqrt(a^2 + r^2 + 2*r*t + t^2)*sqrt(a^2 + r^2 - 2*r*t + t^2)/r^2 + 1/2
```

isolating the radicals :

```
sage: RP=E2.rhs().operands()[2] ; RP
-1/2*sqrt(a^2 + r^2 + 2*r*t + t^2)*sqrt(a^2 + r^2 - 2*r*t + t^2)/r^2
sage: PP=E2.rhs()-RP
sage: E2-PP
u^2 - 1/2*a^2/r^2 - 1/2*t^2/r^2 - 1/2 == -1/2*sqrt(a^2 + r^2 + 2*r*t + t^2)*sqrt(a^2 + r^2 - 2*r*t + t^2)/r^2
```

and squaring again :

```
sage: E4=((E2-PP)^2).expand(); E4
u^4 - a^2*u^2/r^2 - t^2*u^2/r^2 - u^2 + 1/4*a^4/r^4 + 1/2*a^2/r^2 + 1/2*a^2*t^2/r^4 + 1/2*t^2/r^2 + 1/4*t^4/r^4 + 1/4 == 1/4*a^4/r^4 + 1/2*a^2/r^2 + 1/2*a^2*t^2/r^4 - 1/2*t^2/r^2 + 1/4*t^4/r^4 + 1/4
```

Any solution of E is solution of E4, which is now a polynomial equation in `t`

ultimately of the form `A*t^2+B==0`

. In turns out that Maxima can solve it:

```
sage: E4.solve(t)
[t == -sqrt(r^2*u^2/(u^2 - 1) - a^2/(u^2 - 1) - r^2/(u^2 - 1))*u, t == sqrt(r^2*u^2/(u^2 - 1) - a^2/(u^2 - 1) - r^2/(u^2 - 1))*u]
```

$$\left[t = -\sqrt{\frac{r^{2} u^{2}}{u^{2} - 1} - \frac{a^{2}}{u^{2} - 1} - \frac{r^{2}}{u^{2} - 1}} u, t = \sqrt{\frac{r^{2} u^{2}}{u^{2} - 1} - \frac{a^{2}}{u^{2} - 1} - \frac{r^{2}}{u^{2} - 1}} u\right]$$

Since we have squared our equations (twice..), we may have introduced "spurious solutions. The candidate solutions above must be checked. This is not trivial, and is therefore cowardly left to the reader "as an exercise".

It turns out that `sympy`

can solve the original equation E, but that Sage is currently unable to translate it back to Sage:
sage: SS=E.solve(t, algorithm="sympy"); SS

```
ConditionSet(t, Eq(-2*r*u - sqrt(a**2 + (r - t)**2) + sqrt(a**2 + (r + t)**2), 0), {-u*sqrt((-a**2 + r**2*u**2 - r**2)/((u - 1)*(u + 1))), u*sqrt((-a**2 + r**2*u**2 - r**2)/((u - 1)*(u + 1)))})
sage: SS._sage_()
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
<ipython-input-180-8b468472da8b> in <module>()
----> 1 SS._sage_()
AttributeError: 'ConditionSet' object has no attribute '_sage_'
```

One can manually extract the roots proposed by `sympy`

:

```
sage: [u._sage_() for u in SS.args[2].args]
[u*sqrt((r^2*u^2 - a^2 - r^2)/((u + 1)*(u - 1))),
-u*sqrt((r^2*u^2 - a^2 - r^2)/((u + 1)*(u - 1)))]
```

which turn out to be the same as those proposed by Maxima:

$$\left[u \sqrt{\frac{r^{2} u^{2} - a^{2} - r^{2}}{{\left(u + 1\right)} {\left(u - 1\right)}}}, -u \sqrt{\frac{r^{2} u^{2} - a^{2} - r^{2}}{{\left(u + 1\right)} {\left(u - 1\right)}}}\right]$$

**EDIT :** One can also directly call `sympy`

and sk for a dictionary solution simpler to translate back in Sage:

```
sage: import sympy
sage: [{t._sage_():u[t]._sage_() for t in u.keys()} for u in sympy.solve(sympy.sympify(E),
....: t, dict=True)]
[{t: u*sqrt((r^2*u^2 - a^2 - r^2)/(u^2 - 1))},
{t: -u*sqrt((r^2*u^2 - a^2 - r^2)/(u^2 - 1))}]
```

**EDIT 2:** Along the same lines, one can use FriCAS to solve the same equation (with contorsions to to convert the result, again...):

```
sage: SG=fricas.solve(*[fricas(v) for v in [g-u, t]])
sage: [SG[w].lhs().sage()==SG[w].rhs().sage() for w in range(len(SG))]
[t == u*sqrt((r^2*u^2 - a^2 - r^2)/(u^2 - 1)),
t == -u*sqrt((r^2*u^2 - a^2 - r^2)/(u^2 - 1))]
```

**EDIT 3:** Ditto for `giac`

, minus the contorsions:

```
sage: giac.solve(giac(u-g), giac(t)).sage()
[sqrt(r^2*u^4 - a^2*u^2 - 2*r^2*u^2 + a^2 + r^2)*u/(u^2 - 1),
-sqrt(r^2*u^4 - a^2*u^2 - 2*r^2*u^2 + a^2 + r^2)*u/(u^2 - 1)]
```

Mathematica gives the same solutions (again not automagically translatable to Sage):

```
sage: mathematica.Solve(E,t)
{{t -> -(Sqrt[-(a^2*u^2) - r^2*u^2 + r^2*u^4]/Sqrt[-1 + u^2])},
{t -> Sqrt[-(a^2*u^2) - r^2*u^2 + r^2*u^4]/Sqrt[-1 + u^2]}}
```

HTH,

The relation $r^2u^4 - (a^2 + r^2 + t^2)u^2 + t^2 = 0$ holds.

Homework ?

Hint: any solution of

`f(t)=g`

isalsosolution of`f(t)^2=g^2`

(but the converse isnottrue...).@rburing and @Emmanuel Charpentier: I am sorry that I did not make the purpose of this question clear. I have now edited the question to clarify it. Please read particularly the note. It is not to find the inverse function per se by hand, but to coax SageMath to do it.