If $u=\sqrt{a^2{\cos }^2\theta +b^2{\sin }^2\theta }+\sqrt{a^2{\sin }^2\theta +b^2{\cos }^2\theta }$ then the difference between the maximum and minimum of $u^2$ is given by

$u^2=a^2+b^2+2\sqrt{(a^2{\cos }^2\theta +b^2{\sin }^2\theta )(a^2{\sin }^2\theta +b^2{\cos }^2\theta )}=a^2+b^2+$

$2\sqrt{(a^4+b^4){\sin }^2\theta {\cos }^2\theta +a^2{b}^2({\cos }^4\theta +{\sin }^4\theta )}=a^2+b^2+$

$2\sqrt{(a^4+b^4){\sin }^2\theta {\cos }^2\theta +a^2{b}^2(1-2si{n}^2\theta {\cos }^2\theta )}=a^2+b^2+$$2\sqrt{a^2{b}^2+{(a^2-b^2)}^2{\sin }^2\theta {\cos }^2\theta }$

$=a^2+b^2+2\sqrt{a^2{b}^2+\frac{{(a^2-b^2)}^2}{4}{\sin }^{2}2\theta }$

$\therefore$difference =(maximum of $u^2$)

-(minimum of $u^2$)

$=2\sqrt{a^2{b}^2+\frac{(a^2-b^2)}{4}}-2ab$

$=\sqrt{{(a^2+b^2)}^2}-2ab={(a-b)}^2$