JEE Main 201910 Apr, Morning shiftMathematicsMediumMatrices and Determinants245× since 2002If Δ1=∣xsinθcosθ−sinθ−x1cosθ1x∣{\Delta _1} = \left| {\matrix{ x & {\sin \theta } & {\cos \theta } \cr { - \sin \theta } & { - x} & 1 \cr {\cos \theta } & 1 & x \cr } } \right|Δ1=x−sinθcosθsinθ−x1cosθ1x and Δ2=∣xsin2θcos2θ−sin2θ−x1cos2θ1x∣{\Delta _2} = \left| {\matrix{ x & {\sin 2\theta } & {\cos 2\theta } \cr { - \sin 2\theta } & { - x} & 1 \cr {\cos 2\theta } & 1 & x \cr } } \right|Δ2=x−sin2θcos2θsin2θ−x1cos2θ1x, x≠0x \ne 0x=0 ; then for all θ∈(0,π2)\theta \in \left( {0,{\pi \over 2}} \right)θ∈(0,2π) :AOption A. Δ1−Δ2{\Delta _1} - {\Delta _2}Δ1−Δ2 = x (cos 2θ\thetaθ – cos 4θ\thetaθ)BOption B. Δ1+Δ2{\Delta _1} + {\Delta _2}Δ1+Δ2 = - 2x³COption C. Δ1+Δ2{\Delta _1} + {\Delta _2}Δ1+Δ2 = – 2(x³ + x –1)DOption D. Δ1−Δ2{\Delta _1} - {\Delta _2}Δ1−Δ2 = - 2x³CheckSkip · show solution
