Question

The value of $\Delta \, = \, \begin{vmatrix} 1 & sin 3 \theta & sin^{3} ⁡ \theta \\ 2cos ⁡ \theta & sin ⁡ 6 \theta & sin^{3} ⁡ 2 \theta \\ 4cos^{2} ⁡ \theta - 1 & sin ⁡ 9 \theta & sin^{3} ⁡ 3 \theta \end{vmatrix}$ is equal to

• A. $-2$
• B. $-1$
• C. $1$
• D. $0$

Answer 1

Answer: (D)

Multiplying $C_{1}$ by $3 \, sin$ $\theta $ and dividing $\Delta $ by $3 \, sin$ $\theta $ , we get,
$\Delta \, =\frac{1}{3 sin \theta } \, \begin{vmatrix} 3sin⁡\theta & sin⁡3\theta & sin^{3}⁡\theta \\ 3sin⁡2\theta & sin⁡6\theta & sin^{3}⁡2\theta \\ 3sin⁡3\theta & sin⁡9\theta & sin^{3}⁡3\theta \end{vmatrix}$
$\because 3sin \theta \, \left(4 \left(cos\right)^{2} ⁡ \theta - 1\right) = 3 sin ⁡ \theta \left(3 - 4 \left(sin\right)^{2} ⁡ \theta \right) = 3 \, \left(sin ⁡ \theta - 4 \left(sin\right)^{3} ⁡ \theta \right) = 3 sin ⁡ 3 \theta $
Now, applying $C_{1}\leftrightarrow C_{1}-4C_{3}$
$C_{1}$ becomes identical with $C_{2}$
Hence, $\Delta =0$