I don't like the term rationalize here. What you've really done is put the expression in a form where plugging in zero for $h$ makes sense: you can't do this in the first form because you can't divide by zero. That is likely what the textbook author's intent was. I think they created this as a silly pre-exercise to what you'll be doing later so you'll "be prepared" to evaluate limits. Soon you'll do something that looks like
$$
\lim_{h \to 0}\frac{\sqrt{4+h}-2}{h} =
\lim_{h \to 0}\frac{1}{\sqrt{4+h}+2} =
\frac{1}{\sqrt{4+0}+2} = \frac{1}{4}
$$
It's always possible.
You want to multiply top and bottom by $M$ to get that $denominator*M$ has not radical.
As you have figured out: If the denominator is $a + b\sqrt{c}$ you mulitply by the conjugate to get $(a + b\sqrt{c})(a - b\sqrt{c}) = a^2 - b^2*c$.
This will also work with $(\sqrt a + \sqrt b)(\sqrt a - \sqrt b) = a - b$.
So it's the same idea for $a + \sqrt[k] b$. The trick is to realize that $(a + \sqrt[k]b)(a^{k-1} - a^{k-2}\sqrt[k]b + a^{k-3}(\sqrt[k]b)^2-..... \pm a(\sqrt[k]b)^{k-2} \mp (\sqrt[k]b)^{k-1} = a^k \pm b$.
Example: To deradicalize $5 + \sqrt[3]7$ multiply by $5^2 - 5*\sqrt[3]7 + (\sqrt[3]7)^2$ to get $(5 + \sqrt[3]7)(5^2 - 5*\sqrt[3]7 + (\sqrt[3]7)^2) = 5^3 + 5^2\sqrt[3]7 -5^2\sqrt[3]7 - 5*(\sqrt[3]7)^2 + 5*(\sqrt[3]7)^2 + (\sqrt[3]7)^3 = 125 + 7$.
So to deradicalize $(2 + \sqrt 2 + \sqrt[3] 2)$ just deradicalize it term by term.
First let's get rid of the $\sqrt[3]2$ term. So we multiply top and bottom by $(2+\sqrt 2)^2 - (2 +\sqrt 2)*\sqrt[3]2 + (\sqrt[3]2)^2$ to get $(2 + \sqrt 2 + \sqrt[3] 2)*[(2+\sqrt 2)^2 - (2 +\sqrt 2)*\sqrt[3]2 + (\sqrt[3]2)^2] = (2 + \sqrt 2)^3 + 2= 8 + 12 \sqrt 2 + 12\sqrt 2 + 2\sqrt 2 + 2 = 10 + 26\sqrt 2$. Then we multiply that by $10 - 26 \sqrt 2$ to get $(10 + 26\sqrt 2)(10 - 26\sqrt 2) = 100 - 2*26^2$.
So example:
\begin{align} &\frac 1 {2 + \sqrt 2 + \sqrt[3] 2} \\&=
\frac {(2 + \sqrt 2)^2 - (2+\sqrt2)\sqrt[3]2 + \sqrt[3]2^2}{(2+\sqrt 2)^3 + 2}\\&=
\frac {(4 + 4\sqrt 2 + 2) -2\sqrt[3] 2 - \sqrt 2\sqrt[3]2 + \sqrt[3]2^2}{10 + 26\sqrt 2}\\&=
\frac {[(4 + 4\sqrt 2 + 2) -2\sqrt[3] 2 - \sqrt 2\sqrt[3]2 + \sqrt[3]2^2](10 - 26\sqrt{2})}{100 - 2*26^2} \end{align}
Okay... admittedly that is a bear... but it is doable.
Best Answer
You might have an error, check that at the end you should have that rationalazing the denominator you should have that $$\cfrac{1+\sin{x}}{\cos{x}},$$ and the numerator you should have that $$\cfrac{\cos{x}}{1-\sin{x}},$$ wich it´s always the same by $(\cos{x})^2+(\sin{x})^2=1$