The value of Δ = | 1 sin 3 θ sin3 ⁡ θ 2cos ⁡ θ sin ⁡ 6 θ sin3 ⁡ 2 θ 4cos2 ⁡ θ - 1 sin ⁡ 9 θ sin3 ⁡ 3 θ | is equal to

Question

The value of Δ = | 1 sin 3 θ sin3 ⁡ θ 2cos ⁡ θ sin ⁡ 6 θ sin3 ⁡ 2 θ 4cos2 ⁡ θ - 1 sin ⁡ 9 θ sin3 ⁡ 3 θ | is equal to (A) -2 (B) -1 (C) 1 (D) 0

Tardigrade Admin · Accepted Answer

Multiplying C1 by 3 sin θ and dividing Δ by 3 sin θ , we get, Δ =(1/3 sin θ ) | 3sinâ¡θ sinâ¡3θ sin3â¡θ 3sinâ¡2θ sinâ¡6θ sin3â¡2θ 3sinâ¡3θ sinâ¡9θ sin3â¡3θ | ∵ 3sin θ (4 (cos)2 â¡ θ - 1) = 3 sin â¡ θ (3 - 4 (sin)2 â¡ θ ) = 3 (sin â¡ θ - 4 (sin)3 â¡ θ ) = 3 sin â¡ 3 θ Now, applying C1rightarrow C1-4C3 C1 becomes identical with C2 Hence, Δ =0

Q. The value of $Δ = ∣ ∣ 1 2 cos θ 4 co s^{2} θ - 1 s in 3 θ s in 6 θ s in 9 θ s i n^{3} θ s i n^{3} 2 θ s i n^{3} 3 θ ∣ ∣$ is equal to

$- 2$

$- 1$

$1$

$0$

Q. The value of Δ=∣∣​12cos⁡θ4cos2⁡θ−1​sin3θsin⁡6θsin⁡9θ​sin3⁡θsin3⁡2θsin3⁡3θ​∣∣​ is equal to

−2

−1

1

0

Q. The value of $Δ = ∣ ∣ 1 2 cos θ 4 co s^{2} θ - 1 s in 3 θ s in 6 θ s in 9 θ s i n^{3} θ s i n^{3} 2 θ s i n^{3} 3 θ ∣ ∣$ is equal to

$- 2$

$- 1$

$1$

$0$