" Let "Delta=|[1,sin theta,1],[-sin theta,1,sin theta],[-1,-sin theta,1]|" .The "Delta" lies in the interval: "

Let Delta=|(1,sintheta,1),(-sin theta,1,sintheta),(-1,-sintheta,1)| , then Delta lies in the interval.

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

Let Delta=|(1,sin theta, 1),(-sin theta, 1, sin theta),(-1,-sin theta,1)|0,le theta le 2pi . The

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

Let Delta(theta)=|(1, sin theta, 1),(sin theta, 1, sin theta),(-1, -sin theta , 1)|, 0 le theta le 2pi solution of Delta(theta)=3 is

Let A=[[ 1 , sin theta , 1 ],[-sin theta , 1 , sin theta],[-1 , -sin theta , 1 ]] , where 0 le theta le 2 pi , then,

Let A = [(1,sin theta,1),(- sin theta,1,sin theta),(-1,-sin theta,1)], " where " 0 le theta lt 2 pi . then, which of the following is not correct ?

Let A = [(1,sin theta,1),(- sin theta,1,sin theta),(-1,-sin theta,1)], " where " 0 le theta lt 2 pi . then, which of the following is not correct ?