Right answer is (d) \(\frac{6+\sqrt{3}}{33}\)
Best explanation: When the denominator of an expression contains a term with a square root, the process of converting it to an equivalent expression whose denominator is a rational number is called rationalising the denominator.
By multiplying \(\frac{1}{(6-\sqrt{3})}\) by \(6+\sqrt{3}\), we will get same expression since \(\frac{6+\sqrt{3}}{6+\sqrt{3}}\) = 1.
Therefore, \(\frac{1}{(6-\sqrt{3})} = \frac{1}{(6-\sqrt{3})} * \frac{6+\sqrt{3}}{6+\sqrt{3}} = \frac{6+\sqrt{3}}{(6*6) – (\sqrt{3}*\sqrt{3})}\)
= \(\frac{6+\sqrt{3}}{(36-3)}\)
= \(\frac{6+\sqrt{3}}{33}\).