Prove the following

Given,             $a=\frac{3+\sqrt{5}}{2}$            $\ldots$ (i)

Now,               $\frac{1}{a}=\frac{2}{3+\sqrt{5}}=\frac{2}{3+\sqrt{5}} \times \frac{3-\sqrt{5}}{3-\sqrt{5}}$         [multiplying numerator and denominator by $3-\sqrt{5}$ ]

$=\frac{6-2 \sqrt{5}}{3^{2}-(\sqrt{5})^{2}}$           [using identity, $(a-b)(a+b)=a^{2}-b^{2}$ ]

$=\frac{6-2 \sqrt{5}}{9-5}=\frac{6-2 \sqrt{5}}{4}$

$\Rightarrow$               $\frac{1}{a}=\frac{2(3-\sqrt{5})}{4}=\frac{3-\sqrt{5}}{2}$                    ... (ii)

$\therefore$                  $a^{2}+\frac{1}{a^{2}}=a^{2}+\frac{1}{a^{2}}+2-2=\left(a+\frac{1}{a}\right)^{2}-2$            [adding and subtracting 2]

$=\left(\frac{3+\sqrt{5}}{2}+\frac{3-\sqrt{5}}{2}\right)^{2}-2$               [from Eqs. (i) and (ii)]

$=\left(\frac{6}{2}\right)^{2}-2=(3)^{2}-2=9-2=7$