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4-1.Complex numbers
hard
If $\frac{3+i \sin \theta}{4-i \cos \theta}, \theta \in[0,2 \pi],$ is a real number, then an argument of $\sin \theta+\mathrm{i} \cos \theta$ is
A
$-\tan ^{-1}\left(\frac{3}{4}\right)$
B
$\tan ^{-1}\left(\frac{4}{3}\right)$
C
$\pi-\tan ^{-1}\left(\frac{4}{3}\right)$
D
$\pi-\tan ^{-1}\left(\frac{3}{4}\right)$
(JEE MAIN-2020)
Solution
$\frac{3+i \sin \theta}{4-i \cos \theta}$ is a real number
$\Rightarrow 3 \cos \theta+4 \sin \theta=0$
$\Rightarrow \tan \theta=\frac{-3}{4}$
argument of $\sin \theta+i \cos \theta=\pi-\tan ^{-1} \frac{4}{3}$
Standard 11
Mathematics
Similar Questions
Let $z_k=\cos \left(\frac{2 k \pi}{10}\right)+ i \sin \left(\frac{2 k \pi}{10}\right) ; k =1,2, \ldots 9$.
| List $I$ | List $II$ |
| $P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j=1$ | $1.$ True |
| $Q.$ There exists a $k \in\{1,2, \ldots ., 9\}$ such that $z_{1 .} . z=z_k$ has no solution $z$ in the set of complex numbers. | $2.$ False |
| $R.$ $\frac{\left|1-z_1\right|\left|1-z_2\right| \ldots . .\left|1-z_9\right|}{10}$ equals | $3.$ $1$ |
| $S.$ $1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)$ equals | $4.$ $2$ |
Codes: $ \quad P \quad Q \quad R \quad S$
