If $\frac{3+i \sin \theta}{4-i \cos \theta}$, $\theta \in[0,2 \pi]$, is a real number, then an argument of $\sin \theta+i \cos \theta$ is:

(B) $\pi-\tan ^{-1}\left(\frac{4}{3}\right)$

(A) $-\tan ^{-1}\left(\frac{3}{4}\right)$

(C) $\pi-\tan ^{-1}\left(\frac{3}{4}\right)$

(D) $\tan ^{-1}\left(\frac{4}{3}\right)$

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Complex Numbers

\frac{3 + i s i n θ}{4 - i c o s θ}

θ \in [0, 2 π]

, is a real number, then an argument of

sin θ + i cos θ

is:

Answer: (B)

Multiplying numerator and denominator by the conjugate

4 + i cos θ

makes the denominator real, and the imaginary part of the numerator is

4 sin θ + 3 cos θ

. For the quotient to be real this must vanish, so

tan θ = - \frac{3}{4}

. Taking

θ

in the fourth quadrant,

sin θ = - \frac{3}{5}

and

cos θ = \frac{4}{5}

, so

sin θ + i cos θ = - \frac{3}{5} + i \frac{4}{5}

lies in the second quadrant, and its argument is

π - tan^{- 1} \frac{4/5}{3/5} = π - tan^{- 1} \frac{4}{3}

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option