If A = 1 sin θ 1 - sin θ 1 sin θ - 1 - sin θ 1 , then for all θ ∈ 3 π 4 , 5 π 4 , det A lies in the interval :

Step-by-step solution

A= [ array ccc 1 & & 1 - & 1 & -1 & - & 1 array ] det (A)= (11+ ^2 )- (- + )+1 ( ^2 +1 ) =2+2 ^2 ( 3 4 , 5 4 ) 2 [0, 1 2 ] det (A) [2,3] which is a subset of ( 3 2 , 3 ]