`[[1,sintheta,1],[-sintheta,1,sintheta],[-1,-sintheta,1]]show 2<=/_\<=4`

Let A=[[1,sintheta,1],[-sintheta,1,sintheta],[-1,-sintheta,1]] , where 0lt=thetalt=2pi . Then (A) D e t(A) = 0 (B) D e t(A) in (2,oo) (C) D e t(A) in (2, 4) (D) D e t (A) in [2, 4]

Let A=|[1,sintheta,1],[-sintheta,1,sintheta],[-1,-sintheta,1]|, where 0lt=thetalt=2pidot Then

Let A=|[1,sintheta,1],[-sintheta,1,sintheta],[-1,-sintheta,1]|, where 0lt=thetalt=2pi . Then

Let A=|[1,sintheta,1],[-sintheta,1,sintheta],[-1,-sintheta,1]|, where 0lt=thetalt=2pidot Then (a) D e t(A)=0 (b) D e t(A) in (2,oo) (c) D e t(A) in (2,4) (d) D e t(A) in [2,4]

If A = [(1,sintheta,1),(-sintheta,1,sintheta),(-1,-sintheta,1)] , then for all thetain ((3pi)/(4),(5pi)/(4)) , det. (A) lies in the interval

Let A=[(1, sintheta, 1),(-sintheta, 1, sintheta),(-1, -sintheta, 1)], thetaepsilon((3pi)/4, (5pi)/4), then |A| lies in (A) [1/2,3) (B) [1,4) (C) [-1,2] (D) none of these

Let A=|(1,sintheta,1),(-sintheta,1,sintheta),(-1,-sintheta,1)|, where 0lt=thetalt=2pidot Then D e t(A)=0 (b) D e t(A) in (2,oo) D e t(A) in (2,4) (d) D e t(A) in [2,4]

If Delta=|{:(1,sintheta,1),(-sintheta,1,sintheta),(-1,-sintheta,1):}|"prove that" 2lt=Deltalt=4.

If Delta = |[1, 3costheta, 1], [sintheta, 1, 3costheta], [1, sintheta, 1]| , the maximum value of Delta is

Prove that the determinant |[x, sintheta,costheta],[-sintheta,-x,1],[costheta,1,x]| is independent of theta