Trigonometry Value Calculation

We are given the equation $3 \tan \theta = 2$.

From this, we can determine the value of $\tan \theta$: $ \tan \theta = \frac{2}{3} $

Finding Sine and Cosine Values

We can visualize a right-angled triangle where $\theta$ is one of the acute angles. Since $\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{2}{3}$, we can set the length of the opposite side to $2k$ and the adjacent side to $3k$ for some constant $k$. Using the Pythagorean theorem, the hypotenuse $h$ is calculated as: $ h^2 = (\text{opposite})^2 + (\text{adjacent})^2 $ $ h^2 = (2k)^2 + (3k)^2 $ $ h^2 = 4k^2 + 9k^2 $ $ h^2 = 13k^2 $ $ h = \sqrt{13k^2} = k\sqrt{13} $

Now we can find the values of $\sin \theta$ and $\cos \theta$: $ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2k}{k\sqrt{13}} = \frac{2}{\sqrt{13}} $ $ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{3k}{k\sqrt{13}} = \frac{3}{\sqrt{13}} $

Evaluating the Expression

The expression we need to evaluate is: $ \frac{\sqrt{13} \sin \theta - 3 \tan \theta}{3 \tan \theta + \sqrt{13} \cos \theta} $

Substitute the known values: $\tan \theta = \frac{2}{3}$, $\sin \theta = \frac{2}{\sqrt{13}}$, and $\cos \theta = \frac{3}{\sqrt{13}}$. $ \frac{\sqrt{13} \left(\frac{2}{\sqrt{13}}\right) - 3 \left(\frac{2}{3}\right)}{3 \left(\frac{2}{3}\right) + \sqrt{13} \left(\frac{3}{\sqrt{13}}\right)} $

Simplify the numerator and the denominator: Numerator: $ \sqrt{13} \times \frac{2}{\sqrt{13}} - 2 = 2 - 2 = 0 $ Denominator: $ 2 + \sqrt{13} \times \frac{3}{\sqrt{13}} = 2 + 3 = 5 $

Therefore, the value of the expression is: $ \frac{0}{5} = 0 $