If cot x = 3, then what will be the value of $$\frac{(3 + 3 \sin x)(1 - \sin{x})}{(2 + 2 \cos{x})(3 - 3 \cos{x})}$$?​

Correct option is B

Given:

cot x = 3

Formula used:

​$$\begin{aligned}& \mathrm{A}^2-\mathrm{B}^2=(\mathrm{A}+\mathrm{B})(\mathrm{A}-\mathrm{B}) \\& \operatorname{Cot} \theta=\frac{\operatorname{Cos} \theta}{\operatorname{Sin} \theta}\end{aligned}$$

Solution:

​$$\begin{aligned}& \frac{(3+3 \sin \mathrm{x})(1-\sin \mathrm{x})}{(2+2 \cos \mathrm{x})(3-3 \cos \mathrm{x})} \\\Rightarrow & \frac{3 \times(1+\sin x)(1-\sin x)}{6 \times(1+\cos x)(1-\cos x)} \\\Rightarrow & \frac{\left(1-\sin ^2 x\right)}{2 \times\left(1-\cos ^2 x\right)} \\\Rightarrow & \frac{\cos ^2 x}{2 \times \sin ^2 x} \\\Rightarrow & \frac{\cot ^2 x}{2} \\\Rightarrow & \frac{3^2}{2} \\\Rightarrow & \frac{9}{2}\end{aligned}$$ 

Thus, the correct answer is option (b).​