A) \[2-\sqrt{3}\]
B) \[2+\sqrt{3}\]
C) \[16-\sqrt{3}\]
D) \[40-\sqrt{3}\]
Correct Answer: C
Solution :
\[\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt{3}-1}\] \[=\frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})((2+\sqrt{3}))}+\frac{(2-\sqrt{3})(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}\] \[+\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}-1)(\sqrt{3}-1)}\] \[=\frac{{{(2+\sqrt{3})}^{2}}}{4-3}+\frac{(2-\sqrt{3})}{4-3}+\frac{(\sqrt{3}-1)}{3-1}\] \[=4+3+4\sqrt{3}+4+3-4\sqrt{3}+\frac{3+1-2\sqrt{3}}{2}\] \[=14+2-\sqrt{3}\] \[=16-\sqrt{3}\]
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