Solution
a = 2 + √3
$$(a^{2}+\frac{1}{a^{2}})$$ = $$(a + \frac{1}{a})^2$$ - 2
here ,
$$a = 2+\sqrt{3}$$
$$\dfrac{1}{a} = \dfrac{1}{2+\sqrt{3}} \times\dfrac{2-\sqrt{3}}{2-\sqrt{3}} = 2-\sqrt{3}$$
$$a+\dfrac{1}{a} = 2+\sqrt{3} + 2-\sqrt{3}$$
$$a + \dfrac{1}{a} = 4$$
So, $$(a + \frac{1}{a})^2$$ - 2 = 14
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