Correct Answer: 0

Solution :
$2 \cot \theta = 3$
$⇒\cot \theta=\frac{3}{2}$
$⇒\frac{\text{Base}}{\text{Perpendicular}} =\frac{3}{2}$
$⇒\frac{AB}{BC}= \frac{3}{2}$
Let $AB = 3k, BC = 2k$
From Pythagoras' Theorem,
$AC^{2} = AB^{2} + BC^{2}$
$⇒AC^{2} = (3k)^{2} + (2k)^{2}$
$⇒AC^{2} = 13k^2$
$⇒AC = \sqrt{13}k$
$\sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{2k}{\sqrt{13}k}=\frac{2}{\sqrt{13}}$
$\cos\theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{3k}{\sqrt{13}k}= \frac{3}{\sqrt{13}}$
$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{2k}{3k}=\frac{2}{3}$
Putting these values in the equation, we get,
$=\frac{\sqrt{13} \sin \theta – 3 \tan \theta}{3 \tan \theta + \sqrt{13} \cos \theta}$
$=\frac{\sqrt{13} \times \frac{2}{\sqrt{13}} - 3 \times \frac{2}{3}}{3 \times\frac{2}{3} + \sqrt{13} \times \frac{3}{\sqrt{13}}}$
$=0$
Hence, the correct answer is 0.