If `Delta=|(1,3cos theta,1),(sin theta,1,3cos theta),(1,sin theta,1)|,` then the `1/1000` [|maximum value of `Delta-` minimum value of `Delta|^3`] is equal to.

If Delta=det[[1,3cos theta,1sin theta,1,3cos theta1,sin theta,1]] then the (1)/(1000),[maximum value of Delta- minimum value of Delta|^(3)] is equal to

If Delta=|(1, sin theta, 1 ),(-sin theta, 1, sin theta),(-1 , -sin theta, 1)| then Delta lies in the interval

If f(theta)=|(1, cos theta, 1),(-sin theta, 1,-cos theta),(-1,sin theta,1)| and A and B are respectively the maximum and the minimum values of f(theta) , then (A, B) is equal to

let Delta=det[[1,sin theta,1-sin theta,1,sin theta-1,-sin theta,1]],0<=theta<=2 pi

Let A=[(1,3cos 2theta,1),(sin2theta, 1, 3 cos 2 theta),(1, sin 2 theta, 1)] the maximum value of |A| is equal to k, then (k-3)^(2) is equal to

Let A=[(1,3cos 2theta,1),(sin2theta, 1, 3 cos 2 theta),(1, sin 2 theta, 1)] the maximum value of |A| is equal to k, then (k-3)^(2) is equal to

If Delta =|{:(" "1,cos theta,1),(-cos theta," "1,cos theta),(-1,-costheta," "1):}|,then Delta "lies in" the intervail

Let f(theta)=|(1,cos theta, -1),(-sin theta, 1, -cos theta),(-1,sin theta, 1)| Suppose A and B are respectively maximum and minimum value of f(theta) . Then (A,B) is equal to