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Consider the following statements in respect of the determinant $\left|\begin{array}{ll}\cos ^{2} \frac{\alpha}{2} & \sin ^{2} \frac{\alpha}{2} \\ \sin ^{2} \frac{\beta}{2} & \cos ^{2} \frac{\beta}{2}\end{array}\right|$
where $\alpha, \beta$ are complementary angles
$1.$ The value of the determinant is $\frac{1}{\sqrt{2}} \cos \left(\frac{\alpha-\beta}{2}\right)$.
$2.$ The maximum value of the determinant is $\frac{1}{\sqrt{2}}$.
Which of the above statements is/ are correct?
Options:
where $\alpha, \beta$ are complementary angles
$1.$ The value of the determinant is $\frac{1}{\sqrt{2}} \cos \left(\frac{\alpha-\beta}{2}\right)$.
$2.$ The maximum value of the determinant is $\frac{1}{\sqrt{2}}$.
Which of the above statements is/ are correct?
Solution:
1521 Upvotes
Verified Answer
The correct answer is:
Both 1 and 2
$\alpha+\beta=90^{\circ}$
$\left|\begin{array}{rr}\cos ^{2} \frac{\alpha}{2} & \sin ^{2} \frac{\alpha}{2} \\ \sin ^{2} \frac{\beta}{2} & \cos ^{2} \frac{\beta}{2}\end{array}\right|$
$=\cos ^{2} \frac{\alpha}{2} \cos ^{2} \frac{\beta}{2}-\sin ^{2} \frac{\alpha}{2} \sin ^{2} \frac{\beta}{2}$
$=\left(\cos \frac{\alpha}{2} \cos \frac{\beta}{2}+\sin \frac{\alpha}{2} \sin \frac{\beta}{2}\right)\left(\cos \frac{\alpha}{2} \cos \frac{\beta}{2}-\sin \frac{\alpha}{2} \sin \frac{\beta}{2}\right)$
$=\cos \frac{(\alpha-\beta)}{2} \times \cos \frac{(\alpha+\beta)}{2}$
$=\cos \frac{(\alpha-\beta)}{2} \times \cos \frac{\left(90^{\circ}\right)}{2}$
$=\cos \frac{(\alpha-\beta)}{2} \times \frac{1}{\sqrt{2}}$
Maximum value of $\cos \left(\frac{\alpha-\beta}{2}\right)$ is $1 .$ Somaximum value
of determinent is $\left(\frac{1}{\sqrt{2}}\right)$
So both 1 and 2 are correct.
$\left|\begin{array}{rr}\cos ^{2} \frac{\alpha}{2} & \sin ^{2} \frac{\alpha}{2} \\ \sin ^{2} \frac{\beta}{2} & \cos ^{2} \frac{\beta}{2}\end{array}\right|$
$=\cos ^{2} \frac{\alpha}{2} \cos ^{2} \frac{\beta}{2}-\sin ^{2} \frac{\alpha}{2} \sin ^{2} \frac{\beta}{2}$
$=\left(\cos \frac{\alpha}{2} \cos \frac{\beta}{2}+\sin \frac{\alpha}{2} \sin \frac{\beta}{2}\right)\left(\cos \frac{\alpha}{2} \cos \frac{\beta}{2}-\sin \frac{\alpha}{2} \sin \frac{\beta}{2}\right)$
$=\cos \frac{(\alpha-\beta)}{2} \times \cos \frac{(\alpha+\beta)}{2}$
$=\cos \frac{(\alpha-\beta)}{2} \times \cos \frac{\left(90^{\circ}\right)}{2}$
$=\cos \frac{(\alpha-\beta)}{2} \times \frac{1}{\sqrt{2}}$
Maximum value of $\cos \left(\frac{\alpha-\beta}{2}\right)$ is $1 .$ Somaximum value
of determinent is $\left(\frac{1}{\sqrt{2}}\right)$
So both 1 and 2 are correct.
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