If (3+i sinθ/4-i cosθ), θ ϵ [0, 2π], is a real number, then an argument of sinθ + icosθ is :

Question

If (3+i sinθ/4-i cosθ), θ ϵ [0, 2π], is a real number, then an argument of sinθ + icosθ is : (A) π-tan-1((4/3)) (B) -tan-1((3/4)) (C) π-tan-1((3/4)) (D) tan-1((4/3))

Tardigrade Admin · Accepted Answer

(3+i sin θ/4-i cos θ) is a real number ⇒ 3 cos θ +4 sin θ =0 ⇒ tan θ = (-3/4) argument of sin θ +icos θ =π-tan-1 (4/3)

Q. If $\frac{3 + i s in θ}{4 - i cos θ}, θ ϵ [0, 2 π],$ is a real number, then an argument of $s in θ + i cos θ$ is :

$π - t a n^{- 1} (\frac{4}{3})$

$- t a n^{- 1} (\frac{3}{4})$

$π - t a n^{- 1} (\frac{3}{4})$

$t a n^{- 1} (\frac{4}{3})$

$\frac{3 + i s in θ}{4 - i cos θ}$ is a real number
$\Rightarrow 3 cos θ + 4 s in θ = 0$
$\Rightarrow t an θ = \frac{- 3}{4}$
argument of $s in θ + i cos θ = π - t a n^{- 1} \frac{4}{3}$

Q. If 4−icosθ3+isinθ​,θϵ[0,2π], is a real number, then an argument of sinθ+icosθ is :

π−tan−1(34​)

−tan−1(43​)

π−tan−1(43​)

tan−1(34​)