Question:
If $a=2+\sqrt{3}$, then find the value of $\left(a-\frac{1}{a}\right)$.
Solution:
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, $(a+b)(a-b)=a^{2}-b^{2}$ ]
$\Rightarrow \quad \frac{1}{a}=2-\sqrt{3}$
$\therefore \quad a-\frac{1}{a}=(2+\sqrt{3})-(2-\sqrt{3})=2+\sqrt{3}-2+\sqrt{3}=2 \sqrt{3}$
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