JEE Advanced 2023 — Paper 1 — Question 10

$\frac{281(7+6 i \sin \theta)}{7-3 i \cos \theta} \times \frac{7+3 i \cos \theta}{7+3 i \cos \theta}$ $=\frac{281(49-9 \sin 2 \theta)}{49+9 \cos ^{2} \theta}+\frac{5901 i(2 \sin \theta+\cos \theta)}{49+9 \cos ^{2} \theta}$ for it to be positive integer (i.e. real number) $2 \sin \theta+\cos \theta=0$ $\Rightarrow \frac{281(49-9 \sin 2 \theta)}{49+9 \cos ^ {2} \theta}=\frac{281(49+9 \cos ^ {2}\theta)}{49+9 \cos ^{2}\theta}=281$