Q. If $f(\theta) = \left| \begin{matrix} \cos^{2}\theta & \cos{\theta\sin\theta} & - sin\theta \\ \cos{\theta\sin\theta} & \sin^{2}\theta & cos\theta \\ \sin\theta & - \cos\theta & 0 \end{matrix} \right|.\$Then, for all $\theta$

A) $f(\theta) = 0$
B) $f(\theta) = 1$
C) $f(\theta) = - 1$
D) None of these
Explanation

(b)
Given $f(\theta) = \left| \begin{matrix}
\cos^{2}\theta & \cos{\theta\sin\theta} & - sin\theta \\
\cos{\theta\sin\theta} & \sin^{2}\theta & \cos\theta \\
\sin\theta & - \cos\theta & 0
\end{matrix} \right|$
$= \cos^{2}{\theta(0 + \cos^{2}{\theta) - \cos{\theta\sin\theta(0 - \sin{\theta\cos{\theta)}}}}}$
$- \sin{\theta( - \cos^{2}{\theta\sin{\theta - \sin^{3}{\theta)}}}}$
$= \cos^{4}{\theta + 2\sin^{2}{\theta\cos^{2}{\theta + \sin^{4}\theta}}}$
$= \cos^{4}{\theta + \sin^{2}{\theta\cos^{2}{\theta + \sin^{2}\theta}}}$
$= \cos^{2}{\theta(\cos^{2}{\theta + \sin^{2}{\theta) + \sin^{2}{\theta = 1}}}}$
$\therefore For\ all$, $\theta,\ f(\theta) = 1$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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