If a = 2 + sqrt{3},then find the value of a - | vedclass

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(A) Given that $a = 2 + \sqrt{3}$.First,we find the value of $\frac{1}{a}$ by rationalizing the denominator:$\frac{1}{a} = \frac{1}{2 + \sqrt{3}}$Mult

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$2 \sqrt{3}$

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(A) Given that $a = 2 + \sqrt{3}$.First,we find the value of $\frac{1}{a}$ by rationalizing the denominator:$\frac{1}{a} = \frac{1}{2 + \sqrt{3}}$Multiply the numerator and denominator by the conjugate $(2 - \sqrt{3})$:$\frac{1}{a} = \frac{1}{2 + \sqrt{3}} 	imes \frac{2 - \sqrt{3}}{2 - \sqrt{3}}$Using the identity $(x + y)(x - y) = x^2 - y^2$ in the denominator:$\frac{1}{a} = \frac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2}$$\frac{1}{a} = \frac{2 - \sqrt{3}}{4 - 3} = \frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}$Now,calculate $a - \frac{1}{a}$:$a - \frac{1}{a} = (2 + \sqrt{3}) - (2 - \sqrt{3})$$a - \frac{1}{a} = 2 + \sqrt{3} - 2 + \sqrt{3}$$a - \frac{1}{a} = 2\sqrt{3}$

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

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