Question

Choose the correct answer from the alternatives given.
If $a(2+\sqrt{3})=b(2-\sqrt{3})=1,$ then the value of $\frac{1}{a^2+1}+\frac{1}{b^2+1}$ is

Answer 1

$a(2+\sqrt{3})=1$
$\Rightarrow a=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}\Rightarrow a^2=4+3-4\sqrt{3}$
also, $b=\frac{1}{2-\sqrt{3}}=2+\sqrt{3}$
$\Rightarrow b^2=4+3+4\sqrt{3}$
Now $\frac{1}{a^2+1}+\frac{1}{b^2+1}=\frac{1}{7-4\sqrt{3}+1}+\frac{1}{7+4\sqrt{3}+1}$
$=\frac{8+4\sqrt{3}+8-4\sqrt{3}}{(8-4\sqrt{3})(8+4\sqrt{3})}=\frac{16}{64-48}=\frac{16}{16}=1.$