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Let $A=\left\{\frac{1967+1686 i \sin \theta}{7-3 i \cos \theta}: \theta \in \mathbb{R}\right\}$. If $A$ contains exactly one positive integer $n$, then the value of $\mathrm{n}$ is
$281$
$130$
$140$
$145$
Solution
$A=\frac{1967+1686 i \sin \theta}{7-3 i \cos \theta}$
$=\frac{281(7+6 i \sin \theta)}{7-3 i \cos \theta} \times \frac{7+3 i \cos \theta}{7+3 i \cos \theta}$
$=\frac{281(49-18 \sin \theta \cos \theta+i(21 \cos \theta+42 \sin \theta))}{49+9 \cos ^2 \theta}$
for positive integer
$\operatorname{Im}(A)=0$
$21 \cos \theta+42 \sin \theta=0$
$\tan \theta=\frac{-1}{2} ; \sin 2 \theta=\frac{-4}{5}, \cos ^2 \theta=\frac{4}{5}$
$\operatorname{Re}(A)=\frac{281(49-9 \sin 2 \theta)}{49+9 \cos ^2 \theta}$
$=\frac{281\left(49-9 \times \frac{-4}{5}\right)}{49+9 \times \frac{4}{3}}=281 \text { (+ve integer) }$
