Question: Simplify $\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}} + \frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$
My Attempt:
On rationalizing both fractions separately and then adding (since denominator becomes $\sqrt{3}$) I got $$-2\sqrt{6}+2\sqrt{2+\sqrt{3}}+2\sqrt{2-\sqrt{3}}$$
However the given answer is $$\frac{\sqrt{6}}{3}$$
Even more confusing is that on inputting the problem into wolframalpha the solution is given as $$\sqrt{2}$$
I have broken my head over this for a couple of hours and I just can't find a solution. Hope someone can help.
Best Answer
You are right, the correct result for the sum of the two given ratios is $\sqrt{2}$. It is $\frac{\sqrt{6}}{3}$ when you take the difference. Presumably there is a typo in your book.
Revise your work and notice that $$\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2 =2+\sqrt{3}+2\underbrace{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}_1+2-\sqrt{3}=6,$$ and $$\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2 =2+\sqrt{3}-2\underbrace{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}_1+2-\sqrt{3}=2.$$
P.S. As a complement you may take a look at nested radicals. The above computations imply that $$2\sqrt{2\pm\sqrt{3}}=\sqrt{6}\pm\sqrt{2}.$$