If sin 2x = 1, then ∣∣ ∣∣0cosx−sinxsinx0

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If sin 2x = 1, then

{∣ ∣∣ ∣ \begin{matrix} 0 & c o s x & - s i n x s i n x & 0 & c o s x c o s x & s i n x & 0 \end{matrix} ∣ ∣∣ ∣}^{2}

equals

Options:

A . 3

B . 0

C . 1

D . None of these

Answer: Option B
:
B

∴ s i n 2 x = 1 t h e n X = \frac{π}{4}

Then,

{∣ ∣∣ ∣ \begin{matrix} 0 & c o s x & - s i n x s i n x & 0 & c o s x c o s x & s i n x & 0 \end{matrix} ∣ ∣∣ ∣}^{2} = {∣ ∣∣∣∣∣ ∣ \begin{matrix} 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{matrix} ∣ ∣∣∣∣∣ ∣}^{2}

= (\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}) {∣ ∣∣ ∣ \begin{matrix} 0 & 1 & - 1 1 & 0 & 1 1 & 1 & 0 \end{matrix} ∣ ∣∣ ∣}^{2}

= \frac{1}{8} {0 - 1 (0 - 1) - 1 (1)}^{2}

=0

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