• # question_answer If${{\Delta }_{1}}=\left| \begin{matrix} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & l & x \\ \end{matrix} \right|$and ${{\Delta }_{2}}=\left| \begin{matrix} x & \sin 2\theta & \cos 2\theta \\ -\sin 2\theta & -x & 1 \\ \cos 2\theta & \text{l} & x \\ \end{matrix} \right|,x\ne 0;$then for all$\theta \in \left( 0,\frac{\pi }{2} \right):$ [JEE Main 10-4-2019 Morning] A) ${{\Delta }_{1}}-{{\Delta }_{2}}=x(cos2\theta -cos4\theta )$B) ${{\Delta }_{1}}+{{\Delta }_{2}}=-2{{x}^{3}}$C) ${{\Delta }_{1}}-{{\Delta }_{2}}=-2{{x}^{3}}$D) ${{\Delta }_{1}}+{{\Delta }_{2}}=-2({{x}^{3}}+x-1)$

${{\Delta }_{1}}=f(\theta )=\left| \begin{matrix} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & \text{l} \\ \cos \theta & \text{l} & x \\ \end{matrix} \right|=-{{x}^{3}}$             and${{\Delta }_{2}}=f(2\theta )=\left| \begin{matrix} x & \sin 2\theta & \cos 2\theta \\ -\sin 2\theta & -x & \text{l} \\ \cos 2\theta & \text{l} & x \\ \end{matrix} \right|=-{{x}^{3}}$             So${{\Delta }_{1}}+{{\Delta }_{2}}=-2{{x}^{3}}$