If $\Delta_{1}=\left|\begin{array}{ccc}\mathrm{x} & \sin \theta & \cos \theta \\ \sin \theta & -\mathrm{x} & 1 \\ \cos \theta & 1 & \mathrm{x}\end{array}\right|$ and $\Delta_{2}=\left|\begin{array}{ccc}x & \sin 2 \theta & \cos 2 \theta \\ \sin 2 \theta & -x & 1 \\ \cos 2 \theta & 1 & x\end{array}\right|, x \neq 0$; then for all $\theta \in\left(0, \frac{\pi}{2}\right)$:

(D) $\Delta_{1}+\Delta_{2}=-2 x^{3}$

(A) $\Delta_{1}-\Delta_{2}=-2 x^{3}$

(B) $\Delta_{1}+\Delta_{2}=-2\left(\mathrm{x}^{3}+\mathrm{x}-1\right)$

(C) $\Delta_{1}-\Delta_{2}=x(\cos 2 \theta-\cos 4 \theta)$

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Determinants

Δ_{1} = x sin θ cos θ sin θ - x 1 cos θ 1 x

and

Δ_{2} = x sin 2 θ cos 2 θ sin 2 θ - x 1 cos 2 θ 1 x, x \neq = 0

; then
for all

θ \in (0, \frac{π}{2})

Answer: (D)

Expanding

Δ_{1}

along the first row gives

x (- x^{2} - 1) - sin θ (- x sin θ - cos θ) + cos θ (- sin θ + x cos θ)

, and after the cross terms cancel this collapses to

- x^{3} - x + x (sin^{2} θ + cos^{2} θ) = - x^{3}

. The same expansion for

Δ_{2}

, with

2 θ

in place of

θ

, again gives

- x^{3}

. Therefore

Δ_{1} + Δ_{2} = - 2 x^{3}

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option