" Evaluate the determinant "Delta=|[1,sin theta,1],[-sin theta,1,sin theta],[-1,-sin theta,1]|." Also,prove that "2<=Delta<=4

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

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

Prove that the determinant |[x,sin theta,cos theta],[-sin theta,-x,1],[cos theta,1,x]| is independent of

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

If theta, phi epsilon R , then the determinant Delta=|(cos theta, - sin theta, 1),(sin theta, cos theta, 1),(cos(theta+phi),-sin(theta+phi),0)| lies in the interval

Evaluate the following determinants: (b) |(cos theta, -sin theta),(sin theta, cos theta)| = cos theta (cos theta) - sin theta(-sin theta) = cos^(2) theta + sin^(2) theta = 1

If sin theta=1/2 then prove that 3sin theta-4sin^3 theta=1

Prove that (1+sin theta)/(1-sin theta)-(1-sin theta)/(1+sin theta)=4sec theta tan theta