Question

Question 12
If $a=2+\sqrt{3}$ , then find the value of $(a-\frac{1}{a})$.

Answer 1

We have $a=2+\sqrt{3}$
Then,$\frac{1}{a}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}$ [Multiplying numerator and denominator by $(2-\sqrt{3})]$
$\frac{2-\sqrt{3}}{(2{)}^2-(\sqrt{3}{)}^2}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$ [Using identity $(a+b)(a-b)=a^2-b^2]$
$\Rightarrow \frac{1}{a}=2-\sqrt{3}\therefore a-\frac{1}{a}=(2+\sqrt{3})-(2-\sqrt{3})=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}$