Discussion

Explanation:

Let x = $\frac{1}{\sqrt{\sqrt{6+2\sqrt{5}}+\sqrt{5}(7+4\sqrt{5})}}$

Let us first calculate $\sqrt{6+2\sqrt{5}}$ = $\sqrt{5}+1$

Now, x = $\frac{1}{\sqrt{(\sqrt{5}+1)+\sqrt{5}(7+4\sqrt{5})}}$

⇒ x = $\frac{1}{\sqrt{(\sqrt{5}+1)+7\sqrt{5}+20}}$

⇒ x = $\frac{1}{\sqrt{21+8\sqrt{5}}}$

Now, $\sqrt{21+8\sqrt{5}}$ = $4+\sqrt{5}$

⇒ x = $\frac{1}{4+\sqrt{5}}$

On rationalising with 4 - √5, we get

⇒ x = $\frac{4-\sqrt{5}}{11}$

Hence, option (c).

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