A) \[16+\sqrt{3}\]
B) \[4-\sqrt{3}\]
C) \[2-\sqrt{3}\]
D) \[2+\sqrt{3}\]
Correct Answer: A
Solution :
Expression \[=\left( \frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}} \right)\] \[\frac{\sqrt{3}+1}{\sqrt{3}-1}\times \frac{\sqrt{3}-1}{\sqrt{3}+1}\] \[=\left[ \frac{{{\left( 2+\sqrt{3} \right)}^{2}}+{{\left( 2-\sqrt{3} \right)}^{2}}}{\left( 2-\sqrt{3} \right)\left( 2+\sqrt{3} \right)} \right]+\frac{{{\left( \sqrt{3}+1 \right)}^{2}}}{3-1}\] \[=\frac{2(4+3)}{4-3}+\frac{3+1+2\sqrt{3}}{2}\] \[\left[ \begin{align} & \because {{(a+b)}^{2}}+{{(a-b)}^{2}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2({{a}^{2}}+{{b}^{2}}) \\ \end{align} \right]\] \[=14+2+\sqrt{3}=16+\sqrt{3}\]
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