A) \[2\sqrt{3}\]
B) 4
C) 6
D) 8
Correct Answer: B
Solution :
\[\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\] \[=\frac{{{(\sqrt{3}+1)}^{2}}+{{(\sqrt{3}-1)}^{2}}}{(\sqrt{3}-1)(\sqrt{3}+1)}\] \[[{{(a+b)}^{2}}+{{(a-b)}^{2}}=2({{a}^{2}}+{{b}^{2}})]\] \[=\frac{2(3+1)}{3-1}=\frac{2\times 4}{2}=4\]You need to login to perform this action.
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