The number of real numbers λ for which the equality sin ( λ α ) sin α - cos ( λ α ) cos α = λ - 1 , holds for all real α which are not integral multiple of π / 2 is

Step-by-step solution

We have, sin λ α sin α - cos λ α cos α = λ - 1 sin λ α cos α - cos λ α sin α = λ - 1 sin α cos α sin λ - 1 α = λ - 1 sin α cos α sin λ - 1 α = λ - 1 2 sin 2 α It is possible, when λ - 1 = 0 ⇒ λ = 1 and λ - 1 2 = 1 ⇒ λ = 3 ∴   λ has only two solution.