The range of this function are all such that the equation has a solution in the domain i.e. . Now we use the identity that .

This gives,

This becomes,

It order for this to have a solution we require,

But once we have satisfy this inequality we have,

Therefore,

Certainly this is in the domain, but does it solve the equation?

For that we need to substitute but we for we do that note,

This means,

. . . .[*]

Now if then,

would give,

What happens if ?

The answer is sad things happen.

The remaining step is to use[*] to show the solutions do not match upon substitution.

Therefore the range is .