If u = sqrt{a^2 cos^2 theta + b^2 sin^2 theta | vedclass

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(D) Given $u = \sqrt{a^2 \cos^2 	heta + b^2 \sin^2 	heta} + \sqrt{a^2 \sin^2 	heta + b^2 \cos^2 	heta}$.Squaring both sides:$u^2 = (a^2 \cos^2 	h

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$(a - b)^2$

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(D) Given $u = \sqrt{a^2 \cos^2 	heta + b^2 \sin^2 	heta} + \sqrt{a^2 \sin^2 	heta + b^2 \cos^2 	heta}$.Squaring both sides:$u^2 = (a^2 \cos^2 	heta + b^2 \sin^2 	heta) + (a^2 \sin^2 	heta + b^2 \cos^2 	heta) + 2\sqrt{(a^2 \cos^2 	heta + b^2 \sin^2 	heta)(a^2 \sin^2 	heta + b^2 \cos^2 	heta)}$.$u^2 = a^2 + b^2 + 2\sqrt{a^4 \sin^2 	heta \cos^2 	heta + a^2 b^2 \cos^4 	heta + a^2 b^2 \sin^4 	heta + b^4 \sin^2 	heta \cos^2 	heta}$.Using $\sin^2 	heta \cos^2 	heta = \frac{\sin^2 2	heta}{4}$ and $\sin^4 	heta + \cos^4 	heta = 1 - 2\sin^2 	heta \cos^2 	heta = 1 - \frac{\sin^2 2	heta}{2}$:$u^2 = a^2 + b^2 + 2\sqrt{a^2 b^2 (1 - \frac{\sin^2 2	heta}{2}) + (a^4 + b^4) \frac{\sin^2 2	heta}{4}}$.$u^2 = a^2 + b^2 + \sqrt{4a^2 b^2 - 2a^2 b^2 \sin^2 2	heta + (a^4 + b^4) \sin^2 2	heta}$.$u^2 = a^2 + b^2 + \sqrt{4a^2 b^2 + (a^2 - b^2)^2 \sin^2 2	heta}$.For maximum value,$\sin^2 2	heta = 1$: $u^2_{\max} = a^2 + b^2 + \sqrt{4a^2 b^2 + (a^2 - b^2)^2} = a^2 + b^2 + \sqrt{(a^2 + b^2)^2} = 2(a^2 + b^2)$.For minimum value,$\sin^2 2	heta = 0$: $u^2_{\min} = a^2 + b^2 + \sqrt{4a^2 b^2} = a^2 + b^2 + 2ab = (a + b)^2$.The difference is $u^2_{\max} - u^2_{\min} = 2a^2 + 2b^2 - (a^2 + b^2 + 2ab) = a^2 + b^2 - 2ab = (a - b)^2$.

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 what is the difference between the maximum and minimum values of $u^2$?

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