| * 20 c 1 + ^2 & ^2 & ^2 ^2 & 1 + ^2 & ^2 4 4 & 4 4 & 1 + 4 4 | … - Vidyadip

MCQ

$\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}} \right| = 0$ then $\sin \,4\theta $ equal to

A
$1/2$
B
$1$
✓
$-1/2$
D
$-1$

✓

Answer

Correct option: C.

$-1/2$

c
(c) $\left| {\,\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}\,} \right| = 0$

Using ${C_1} \to {C_1} - {C_2},{C_2} \to {C_2} - {C_3}$

==> $\left| {\,\begin{array}{*{20}{c}}1&0&{{{\sin }^2}\theta }\\{ - 1}&1&{{{\cos }^2}\theta }\\0&{ - 1}&{1 + 4\sin 4\theta }\end{array}\,} \right| = 0$

==> $2\,(1 + 2\sin 4\theta ) = 0 \Rightarrow \sin 4\theta = \frac{{ - 1}}{2}$.

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