If $$a^{2}+b^{2}$$+$$\frac{1}{a^{2}}$$+$$\frac{1}{b^{2}}$$=4 then the value of $$a^{2}+b^{2}$$ will be

$$a^2+b^2+\ \frac{\ 1}{a^2}+\ \frac{\ 1}{b^2}=4$$

$$a^2+b^2+\ \frac{\ 1}{a^2}+\ \frac{\ 1}{b^2}-4=0$$

$$a^2-2+\frac{\ 1}{a^2}+b^2-2+\ \frac{\ 1}{b^2}=0$$

$$\left(a-\ \frac{\ 1}{a}\right)^2+\left(b-\ \frac{\ 1}{b}\right)^2=0$$

Sum of squares is zero so both the terms are zero

$$=$$>    $$\left(a-\ \frac{\ 1}{a}\right)^2=0$$,     $$\left(b-\ \frac{\ 1}{b}\right)^2=0$$

$$=$$>         $$a-\ \frac{\ 1}{a}=0$$,            $$b-\ \frac{\ 1}{b}=0$$

$$=$$>                  $$a=\frac{\ 1}{a}$$,                     $$b=\frac{\ 1}{b}$$

$$=$$>                 $$a^2=1$$,                     $$b^2=1$$

Therefore  $$a^2+b^2=1+1$$ = 2