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

If 3+isinθ4-icosθ,θ∈0,2π, is a real number, then an argument of sinθ+icosθ is
  1. π-tan-143
  2. π-tan-134
  3. -tan-134
  4. tan-143

Solution

z=3+isinθ4-icosθ×4+icosθ4+icosθ

Im(z)=3cosθ+4sinθ16+cos2θ
As z is purely real ⇒3cosθ+4sinθ=0⇒tanθ=-34
Argsinθ+icosθ=π+tan-1cosθsinθ=π+tan-1-43=π-tan-143

Asked in: JEE Main 2020 (07 Jan Shift 2)

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