(3+isinθ)(4-icosθ),θ∈(0,2π),is a real number. Then, arg(cosθ+isinθ)=
π-tan-143
-tan-134
π-tan-134
tan-143
Modify the denominator so that it doesn't contain imaginary part.
Let,
z=(3+isinθ)(4-icosθ)=(3+isinθ)(4-icosθ)×(4+icosθ)(4+icosθ)=(12-sinθcosθ+i(4sinθ+3cosθ))(16+cos2θ)
As Z is real, therefore,
(4sinθ+3cosθ)=0⇒tanθ=-34[90°<θ<180°]∴arg[sinθ+icosθ]=π+tan-1cosθsinθ=π-tan-143(∵cotθ=-43)
Therefore, Option(A) is the correct answer..
If 4tanθ=3, then 4sinθ-cosθ4sinθ+cosθ is equal to: