$$\frac{1 - \tan A}{1 + \tan A} = \frac{\tan 3^\circ \tan 15^\circ \tan 30^\circ \tan 75^\circ \tan 87^\circ}{\tan 27^\circ \tan 39^\circ \tan 51^\circ \tan 60^\circ \tan 63^\circ}$$, then the value of $$\cot A$$ is :

$$\frac{1-\tan A}{1+\tan A}=\frac{\tan3^{\circ}\tan15^{\circ}\tan30^{\circ}\tan75^{\circ}\tan87^{\circ}}{\tan27^{\circ}\tan39^{\circ}\tan51^{\circ}\tan60^{\circ}\tan63^{\circ}}$$

$$=$$>  $$\frac{1-\tan A}{1+\tan A}=\frac{\tan3^{\circ}\tan15^{\circ}\tan30^{\circ}\tan\left(90-15\right)^{\circ}\tan\left(90-3\right)^{\circ}}{\tan27^{\circ}\tan39^{\circ}\tan\left(90-39\right)^{\circ}\tan60^{\circ}\tan\left(90-27\right)^{\circ}}$$

$$=$$>  $$\frac{1-\tan A}{1+\tan A}=\frac{\tan3^{\circ}\tan15^{\circ}\tan30^{\circ}\cot15^{\circ}\cot3^{\circ}}{\tan27^{\circ}\tan39^{\circ}\cot39^{\circ}\tan60^{\circ}\cot27^{\circ}}$$

$$=$$>  $$\frac{1-\tan A}{1+\tan A}=\frac{\tan30^{\circ}}{\tan60^{\circ}}$$

$$=$$>  $$\frac{1-\tan A}{1+\tan A}=\frac{\frac{1}{\sqrt{3}}}{\sqrt{3}}$$

$$=$$>  $$\frac{1-\tan A}{1+\tan A}=\frac{1}{3}$$

$$=$$>  $$3-3\tan A=1+\tan A$$

$$=$$>  $$4\tan A=2$$

$$=$$>  $$\tan A=\frac{1}{2}$$

$$=$$>  $$\cot A=2$$

Hence, the correct answer is Option A