MCQ
$\begin{vmatrix}1+\sin^2\theta&\sin^2\theta&\sin^2\theta\\\cos^2\theta&1+\cos^2\theta&\cos^2\theta\\4\sin4 \theta&4\sin4\theta&1+4\sin4\theta\end{vmatrix}=0,$ તો $\sin 4\theta = .............$
- A$\frac{1}{2}$
- B$1$
- ✓$ - \frac{1}{2}$
- D$ - 1$
$\begin{vmatrix}1 & 0 & sin^2\theta \\-1 & 1 & cos^2\theta \\0 & -1 & 1+4sin4\theta\end{vmatrix}=0$
$\because C_{21}(-1),C_{32}(-1))$
$\therefore 1(cos^2\theta+sin^2\theta)+1+4sin4\theta =0$
$($ ત્રીજી હારને અનુલક્ષીને વિસ્તરણ $)$
$\therefore 2+4sin4\theta=0\\\therefore sin 4 \theta =-\frac{1}{2}$
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