If $$\cot \theta = \frac{15}{8}, \theta$$ is an acute angle, then find the value of $$\frac{(1 - \cos \theta)(2 + 2 \cos \theta)}{(2 - 2 \sin \theta)(1 + \sin \theta)}$$.

$$\cot\theta=\frac{15}{8}$$

$$\tan\theta=\frac{8}{15}$$

$$\sec\theta=\sqrt{1+\tan^2\theta\ }=\sqrt{1+\left(\frac{8}{15}\right)^2}=\sqrt{\frac{289}{225}\ }=\frac{17}{15}$$

$$\cos\theta=\frac{15}{17}$$

$$\sin\theta=\sqrt{1-\cos^2\theta\ }=\sqrt{1-\left(\frac{15}{17}\right)^2}=\sqrt{1-\frac{225}{289}}=\sqrt{\frac{64}{289}}=\frac{8}{17}$$

$$\frac{(1-\cos\theta)(2+2\cos\theta)}{(2-2\sin\theta)(1+\sin\theta)}=\frac{\left(1-\frac{15}{17}\right)\left[2+2\left(\frac{15}{17}\right)\right]}{\left[2-2\left(\frac{8}{17}\right)\right]\left(1+\frac{8}{17}\right)}$$

$$=\frac{\left(\frac{2}{17}\right)\left[\frac{64}{17}\right]}{\left[\frac{18}{17}\right]\left(\frac{25}{17}\right)}$$

$$=\frac{2\times64}{18\times25}$$

$$=\frac{64}{225}$$

Hence, the correct answer is Option B