Question : If $x=\sqrt{1+\frac{\sqrt{3}}{2}}-\sqrt{1-\frac{\sqrt{3}}{2}}$, then the value of $\frac{\sqrt{3}-x}{\sqrt{3}+x}$ (corrected to two decimal places) is:
Option 1: 0.25
Option 2: 0.17
Option 3: 0.19
Option 4: 0.27
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Correct Answer: 0.27
Solution :
$x=\sqrt{1+\frac{\sqrt{3}}{2}}-\sqrt{1-\frac{\sqrt{3}}{2}}$
Squaring both sides, we get,
$⇒x^2=1+\frac{\sqrt{3}}{2}+1-\frac{\sqrt{3}}{2}-2(\sqrt{1-\frac{3}{4}})$
$⇒x^2=1$
$x=1$ or $x=-1$
Taking $x=1$, we get,
$⇒\frac{\sqrt{3}-x}{\sqrt{3}+x}$
$=\frac{\sqrt{3}-1}{\sqrt{3}+1}$
$=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times\frac{\sqrt{3}-1}{\sqrt{3}+1}$
$=\frac{(\sqrt{3}-1)^2}{3-1}$
$=2-\sqrt{3}$
$=2-1.73$ [As the value of $\sqrt3=1.73$]
$=0.27$
Hence, the correct answer is 0.27.
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