1.If Δ1=∣xsinθcosθ−sinθ−x1cosθ1x∣\Delta_1 = \begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}Δ1=x−sinθcosθsinθ−x1cosθ1x and Δ2=∣xsin2θcos2θ−sin2θ−x1cos2θ1x∣\Delta_2 = \begin{vmatrix} x & \sin 2\theta & \cos 2\theta \\ -\sin 2\theta & -x & 1 \\ \cos 2\theta & 1 & x \end{vmatrix}Δ2=x−sin2θcos2θsin2θ−x1cos2θ1x, x≠0x \neq 0x=0; then for all θ∈(0,π2)\theta \in \left(0, \dfrac{\pi}{2}\right)θ∈(0,2π):a.Δ1−Δ2=x(cos2θ−cos4θ)\Delta_1 - \Delta_2 = x(\cos 2\theta - \cos 4\theta)Δ1−Δ2=x(cos2θ−cos4θ)b.Δ1+Δ2=−2x3\Delta_1 + \Delta_2 = -2x^3Δ1+Δ2=−2x3c.Δ1−Δ2=−2x3\Delta_1 - \Delta_2 = -2x^3Δ1−Δ2=−2x3d.Δ1+Δ2=−2(x3+x−1)\Delta_1 + \Delta_2 = -2(x^3 + x - 1)Δ1+Δ2=−2(x3+x−1)Login to continueOnly logged in users canattempt or see the solution.
