Question

If sin 2x = 1, then ${\left|\begin{array}{ccc}0& cosx& -sinx\\ sinx& 0& cosx\\ cosx& sinx& 0\end{array}\right|}^2$ equals

• A. 3
• B. \N
• C. 1
• D. None of these

Answer 1

The correct option is B \N

$\therefore sin2x=1thenX=\frac{\pi }{4}$
Then, ${\left|\begin{array}{ccc}0& cosx& -sinx\\ sinx& 0& cosx\\ cosx& sinx& 0\end{array}\right|}^2={\left|\begin{array}{ccc}0& \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\end{array}\right|}^2$
$=(\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}){\left|\begin{array}{ccc}0& 1& -1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right|}^2$
$=\frac{1}{8}{\{0-1(0-1)-1(1\left)\right\}}^2$
=0