Question

Consider the following statements in respect of the determinant $\left|\begin{array}{cc}{\cos }^2\frac{\alpha }{2}& {\sin }^2\frac{\alpha }{2}\\ {\sin }^2\frac{\beta }{2}& {\cos }^2\frac{\beta }{2}\end{array}\right|$ where $\alpha ,\beta$ are complementary angles
1. The value of the determinant is $\frac{1}{\sqrt{2}}\cos \left(\begin{array}{c}\frac{\alpha -\beta }{2}\end{array}\right)$.
2. The maximum value of the determinant is $\frac{1}{\sqrt{2}}$.
Which of the above statements is/are correct?

• A. 1 only
• B. 2 only
• C. Both 1 and 2
• D. Neither 1 nor 2

Answer 1

The correct option is C Both 1 and 2

Solution:

We have, $△=\left[\begin{array}{cc}{\cos }^2\frac{\alpha }{2}& {\sin }^2\frac{\alpha }{2}\\ {\sin }^2\frac{\beta }{2}& {\cos }^2\frac{\beta }{2}\end{array}\right]$

$={\cos }^2\frac{\alpha }{2}{\cos }^2\frac{\beta }{2}-{\sin }^2\frac{\alpha }{2}{\sin }^2\frac{\beta }{2}$

$=(\cos \frac{\alpha }{2}\cos \frac{\beta }{2}+\sin \frac{\alpha }{2}\sin \frac{\beta }{2})$ $(\cos \frac{\alpha }{2}\cos \frac{\beta }{2}-\sin \frac{\alpha }{2}\sin \frac{\beta }{2})$

$=\cos \left(\frac{\alpha -\beta }{2}\right)\cos {45}^{\circ }$ ........ $[\alpha ,\beta ]$ are complementry angles.

$=\frac{1}{\sqrt{2}}\cos \left(\frac{\alpha -\beta }{2}\right)$

Maximum value of $\cos \left(\frac{\alpha -\beta }{2}\right)=1$

So, maximum value of $\frac{1}{\sqrt{2}}\cos \left(\frac{\alpha -\beta }{2}\right)=\frac{1}{\sqrt{2}}$

Hence, C is the correct option.