" Let "A=[[1,sin theta,1],[-sin theta,1,sin theta],[-1,-sin theta,1]]," where "0<=theta<=2 pi" .Then "

Let A=det[[1,sin theta,1-sin theta,1,sin theta-1,-sin theta,1]], where 0<=theta<=2 pi. Then Det(A)=0(b)Det(A)in(2,oo)Det(A)in(2,4)(d) Det (A)in[2,4]

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

Let A= |{:(" "1," "sin theta,1),(-sin theta," "1,sintheta),(-1,-sintheta,1):}| , where 0lethetale2 pi , then

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

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

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

(sin theta)/(1-cos theta)=(1+cos theta)/(sin theta)

cos theta/(1+sin theta)=(1-sin theta)/cos theta

(1-cos theta)/(sin theta)=(sin theta)/(1+cos theta)