[,1,sin theta,1],[ctg A=,sin theta,1,sin theta],[-1,-sin theta,1]]" ,"bar(u)

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, sin theta ),( -1 ,-sin theta , 1 )]:} ,"where" 0 letheta le 2pi , Then

Let A={:[( 1,sin theta , 1),( -sin theta , 1, sin theta ),( -1 ,-sin theta , 1 )]:} ,"where" 0 letheta le 2pi , Then

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

sqrt((1+sin theta)/(1-sin theta)?)?

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

If A= [[cos theta,-sin theta],[sin theta,cos theta]] ,then Adj A is (a) [[cos theta,-sin theta],[cos theta,sin theta]] (b) [[1,0],[0,1]] (c) [[cos theta,sin theta],[-sin theta,cos theta]] (d) [[-1,0],[0,-1]]