# If $\cos 2\theta=0$, then ${\begin{vmatrix}0 & \cos\theta & \sin\theta\\\cos\theta & \sin\theta & 0\\\sin\theta & 0 & \cos\theta\end{vmatrix}}^2$ =

$\begin{array}{1 1} \frac{1}{2} \\ \frac{-1}{2} \\ \pm\frac{1}{2} \\ 0\end{array}$

Toolbox:
• The product of two determinants are defined when both the determinants are of same order.
• If $\Delta_1$ and $\Delta_2$ are two determinants of order 3 then their product is also of order 3.
• The product of two determinants can be obtained by multiplying rows and columns.
${\begin{vmatrix}0 & \cos\theta & \sin\theta\\\cos\theta & \sin\theta & 0\\\sin\theta & 0 & \cos\theta\end{vmatrix}}^2=\begin{vmatrix}0 & \cos\theta & \sin\theta\\\cos\theta & \sin\theta & 0\\\sin\theta & 0 & \cos\theta\end{vmatrix}\begin{vmatrix}0 & \cos\theta & \sin\theta\\\cos\theta & \sin\theta & 0\\\sin\theta & 0 & \cos\theta\end{vmatrix}$
On multiplying rows and columns we get,
$\quad=\begin{vmatrix}0+cos^2\theta+sin^2\theta & 0+sin\theta\cos\theta+0 & 0+0\sin\theta cos\theta\\0+sin\theta cos\theta+0 & cos^2\theta+\sin^2\theta+0 & sin\theta cos\theta+0+0\\0+0+sin\theta cos\theta & sin\theta cos\theta+0+0& sin^2\theta+0+\cos^2\theta\end{vmatrix}$
But we know $sin^2\theta+cos^2\theta=1$ and $sin\theta cos\theta=\frac{1}{2}sin2\theta$
$\quad=\begin{vmatrix}1 & \frac{1}{2}sin2\theta & \frac{1}{2}sin2\theta\\\frac{1}{2}sin2\theta& 1&\frac{1}{2}sin2\theta\\\frac{1}{2}sin2\theta&\frac{1}{2}sin2\theta&1\end{vmatrix}$
It is given $cos2\theta=0\Rightarrow 2\theta=\frac{\pi}{2}$ or $\theta=\frac{\pi}{4}$
But we know $sin\frac{\pi}{2}=1$
$\frac{1}{2}sin\frac{\pi}{2}=\frac{1}{2}$
Substituting we get,
$\Delta=\begin{vmatrix}1 & \frac{1}{2} &\frac{1}{2}\\\frac{1}{2} & 1&\frac{1}{2}\\\frac{1}{2}&\frac{1}{2}&1\end{vmatrix}$
Expanding along $R_1$ we get,
$\Delta=1(1-\frac{1}{4})-\frac{1}{2}(\frac{1}{2}-\frac{1}{4})+\frac{1}{2}(\frac{1}{4}-\frac{1}{2})$
$\Delta=\frac{3}{4}-\frac{1}{2}(\frac{1}{4})+\frac{1}{2}(\frac{-1}{4})$
$\Delta=\frac{3}{4}-\frac{1}{8}-\frac{1}{8}$
$\quad=\frac{3}{4}-\frac{2}{8}$
$\quad=\frac{3}{4}-\frac{1}{4}$
$\quad=\frac{1}{2}$
Hence the value of $\Delta=\frac{1}{2}$