If $$\frac{\sin^2 \theta}{\tan^2 \theta - \sin^2 \theta} = 5, \theta$$ is an acute angle, then the value of $$\frac{24\sin^2\theta-15\sec^2\theta}{6\operatorname{cosec}^2\theta-7\cot^2\theta}$$ is:

$$\frac{\sin^2\theta}{\tan^2\theta-\sin^2\theta}=5$$

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

$$\frac{\sin^2\theta}{\sin^2\theta\ \left(\frac{1\ }{\cos^2\theta\ }-1\right)}=5$$

$$\frac{\cos^2\theta}{1-\cos^2\theta\ }=5$$

$$\frac{\cos^2\theta}{\sin^2\theta\ \ }=5$$

$$\cot^2\theta\ \ =5$$

$$\operatorname{cosec}^2\theta\ =1+\cot^2\theta\ =1+5=6$$

$$\sin^2\theta=\frac{1}{\operatorname{cosec}^2\theta}\ =\frac{1}{6}$$

$$\cos^2\theta\ =1-\sin^2\theta=1-\frac{1}{6}=\frac{5}{6}$$

$$\sec^2\theta\ =\frac{1}{\cos^2\theta\ }=\frac{6}{5}$$

$$\frac{24\sin^2\theta-15\sec^2\theta}{6\operatorname{cosec}^2\theta-7\cot^2\theta}=\frac{24\left(\frac{1}{6}\right)-15\left(\frac{6}{5}\right)}{6\left(6\right)-7\left(5\right)}$$

$$=\frac{4-18}{36-35}$$

$$=-14$$

Hence, the correct answer is Option C