Let A = cos α - sin α sin α cos α , a ∈ R such that A 32 = 0 - 1 1 0 . Then, a value of α is:

Step-by-step solution

∵   A = cos α - sin α sin α cos α   ⇒ A 2 = cos 2 α - sin 2 α sin 2 α cos 2 α ⇒ A 3   = cos 3 α - sin 3 α sin 3 α cos 3 α Similarly, A 32   = cos 32 α - sin 32 α sin 32 α cos 32 α Given, A 32   = 0 - 1 1 0 ⇒   cos 32 α - sin 32 α sin 32 α cos 32 α = 0 - 1 1 0 On comparing cos 32 α = 0 & sin 32 α = 1 , we get, α = π 64