6. If ∆ = 1 3 cos θ 1 sin θ 1 3 cos θ 1 sin θ 1 , the maximum value of ∆ is a ) - 10 b ) - 10 C ) 10 D ) 10 Share with your friends Share 0 Aarushi Mishra answered this ∆=13 cos θ1sin θ13 cos θ1sin θ1Expand along R1∆=1.1-3 cos θ sin θ-3 cos θsin θ-3 cos θ+1.sin2 θ-1∆=1-3 cos θ sin θ-3 cos θ sin θ+9 cos2 θ+sin2 θ-1∆=-6 cos θ sin θ+8 cos2 θ+cos2 θ+sin2 θ∆=-6 cos θ sin θ+8 cos2 θ+1∆=-3×2cos θ sin θ+4×2cos2 θ+1Since 2cos2 θ=1+cos 2θ and 2cos θ sin θ=sin 2θ∆=-3×sin 2θ+4×1+cos 2θ+1∆=-3sin 2θ+4cos 2θ+1+4∆=-3sin 2θ+4cos 2θ+5Note: ∆ will be maximum when -3sin 2θ+4cos 2θ is maximumWe know maximum value of Asin x+Bcos x is A2+B2 max∆=-32+42+5max∆=25+5max∆=10 -1 View Full Answer