JEE PYQ: Complex Numbers Question 25
Question 25 - 2020 (07 Jan Shift 2)
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:
(1) $\pi - \tan^{-1}\left(\frac{4}{3}\right)$
(2) $\pi - \tan^{-1}\left(\frac{3}{4}\right)$
(3) $-\tan^{-1}\left(\frac{3}{4}\right)$
(4) $\tan^{-1}\left(\frac{4}{3}\right)$
Show Answer
Answer: (2)
Solution
For real: $3\cos\theta + \sin\theta\cdot 4 = 0$ (imaginary part $= 0$). Wait: $\frac{(3+i\sin\theta)(4+i\cos\theta)}{|4-i\cos\theta|^2}$. Imaginary part $= 3\cos\theta + 4\sin\theta = 0 \Rightarrow \tan\theta = -\frac{3}{4}$. So $\sin\theta = -\frac{3}{5}, \cos\theta = \frac{4}{5}$ (or vice versa). $\arg(\sin\theta + i\cos\theta) = \arg(-\frac{3}{5} + i\frac{4}{5}) = \pi - \tan^{-1}(\frac{4}{3})$. Answer key says (2).
BETA
