If $\tan \theta=\frac{2}{3}$, what is the value of $\frac{15 \sin ^{2} \theta-3 \cos ^{2} \theta}{5 \sin ^{2} \theta+3 \cos ^{2} \theta} ?$

It is given, $\tan \theta=\frac{2}{3}$

Expression $=\frac{15 \sin ^{2} \theta-3 \cos ^{2} \theta}{5 \sin ^{2} \theta+3 \cos ^{2} \theta}$

$=\frac{\frac{15 \sin ^{2} \theta}{\cos ^{2} \theta}-\frac{3 \cos ^{2} \theta}{\cos ^{2} \theta}}{\frac{5 \sin ^{2} \theta}{\cos ^{2} \theta}+\frac{3 \cos ^{2} \theta}{\cos ^{2} \theta}}$

On dividing the Nʳ and Dʳ by cos² θ,

$=\frac{15 \tan ^{2} \theta-3}{5 \tan ^{2} \theta+3}=\frac{15 \times \frac{4}{9}-3}{5 \times \frac{4}{9}+3}=\frac{60-27}{20+27}=\frac{33}{47}$