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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 values of $u^{2}$ is given by :

Option 1)
$a ^ 2 + b^ 2$

Option 2)
$a ^ 2 - b^ 2$

Option 3)
$(a+b)^2$

Option 4)
$(a-b)^2$

Answers (1)

best_answer

As we have learned

Trigonometric Identities -

$\sin ^{2}\Theta + \cos ^{2}\Theta = 1$

$1 + \tan ^{2}\Theta = \sec ^{2}\Theta$

$1 + \cot ^{2}\Theta = cosec ^{2}\Theta$

- wherein

They are true for all real values of $\Theta$

Double Angle Formula -

Double angle formula

- wherein

These are formulae for double angles.

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

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

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

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

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

$Min (u^2) = a^ 2 + b^2 + 2\sqrt{ a^2 b^2 } = (a+b)^2$

$Max (u^2) = a^ 2 + b^2 + 2\sqrt{\frac{( a^2 b^2)^2}{4} } = 2(a^2+b^2 )$

Diffrence $(a- b)^2$

Option 1)

$a ^ 2 + b^ 2$

Option 2)

$a ^ 2 - b^ 2$

Option 3)

$(a+b)^2$

Option 4)

$(a-b)^2$

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Himanshu

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