If 3 + i s i n θ 4 - i c o s θ , θ ∈ 0 ,2 π , is a real number, then an argument of s i n θ + i c o s θ is

Step-by-step solution

z = 3 + i sin θ 4 - i cos θ × 4 + i cos θ 4 + i cos θ Im ( z ) = 3 cos θ + 4 sin θ 16 + cos 2 θ As z is purely real ⇒ 3 cos θ + 4 sin θ = 0 ⇒ tan θ = - 3 4 A r g sin θ + i cos θ = π + tan - 1 cos θ sin θ = π + tan - 1 - 4 3 = π - tan - 1 4 3