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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$(a - b)^2$

Vedclass · Answer

(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 $u^2$?

Similar Questions

If $A+C=2B$,then $\frac{\cos C-\cos A}{\sin A-\sin C}$ is equal to

In triangle $ABC$,if $\cos A \cos B + \sin A \sin B \sin C = 1$,then $\sin A + \sin B + \sin C =$

If $x+y=\frac{\pi}{2}$,then the maximum value of $\sin x \cdot \sin y$ is

If $A>0, B>0$ and $A+B=\frac{\pi}{3}$,then the maximum value of $\tan A \tan B$ is

The minimum value of $3\sin \theta + 4\cos \theta $ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module