If $5 \cot \theta=3$, find the value of $\frac{6 \sin \theta-3 \cos \theta}{7 \sin \theta+3 \cos \theta}$.

$\frac{11}{40}$ $\frac{20}{41}$ $\frac{21}{44}$ $\frac{44}{21}$

$\frac{21}{44}$

5 cotθ = 3 cotθ = \(\frac{3}{5}\) = \(\frac{B}{P}\)  Now, \(\frac{6sinθ - 3cosθ}{7sinθ + 3cosθ}\) = \(\frac{6P - 3B}{7P + 3B}\)  = \(\frac{6 × 5 - 3×3}{7×5 + 3×3}\) = \(\frac{30 - 9}{35 + 9}\) = \(\frac{21}{44}\)