Question

if $x=\frac{1}{2+\sqrt{3}},y=\frac{1}{2-\sqrt{3}}$ then the value of $\frac{1}{x+1}+\frac{1}{y+1}is:$

• A. $\frac{1}{2}$
• B. $\sqrt{3}$
• C. 1
• D. $\frac{1}{\sqrt{3}}$

Answer 1

Answer: (A)

$\frac{1}{2}$

$x=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$

$x+1=3-\sqrt{3}$

$y=\frac{1}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}$

$y+1=3+\sqrt{3}$

Now

$\frac{1}{x+1}+\frac{1}{y+1}=\frac{1}{3-\sqrt{3}}+\frac{1}{3+\sqrt{3}}$

$=\frac{3+\sqrt{3}+3-\sqrt{3}}{9-3}$

$=\frac{6}{6}=1$.