Question

If $f(\theta)=\begin{vmatrix}(\cos )^2 \theta & \cos \theta \sin \theta & -\sin \theta \\ \cos \theta \sin \theta & (\sin )^2 \theta & \cos \theta \\ \sin \theta & -\cos \theta & 0\end{vmatrix}$ then, $f\left(\frac{\pi }{6}\right)+f\left(\frac{\pi }{3}\right)+f\left(\frac{\pi }{2}\right)+f\left(\frac{2 \pi }{3}\right)+f\left(\frac{5 \pi }{6}\right)+f\left(\pi \right)+.\ldots \ldots .+f\left(\frac{53 \pi }{6}\right)$ is equal to

Answer 1

$f(\theta)=\cos ^{2} \theta\left(\cos ^{2} \theta\right)-\cos \theta \sin \theta(-\cos \theta \sin \theta)-\sin \theta\left(-\cos ^{2} \theta \sin \theta-\sin ^{3} \theta\right)$
$=\cos ^{4} \theta+\sin ^{4} \theta+2 \cos ^{2} \theta \sin ^{2} \theta=\left(\cos ^{2} \theta+\sin ^{2} \theta\right)^{2}=1$
$f\left(\frac{\pi}{6}\right)=f\left(\frac{2 \pi}{6}\right)=f\left(\frac{3 \pi}{6}\right)=\ldots \ldots .=f\left(\frac{53 \pi}{6}\right)=1$