Let $$x = \frac{\pi}{40}$$. Then the value of  $$\cot x \cot 2x \cot 3x ....\cot 19x $$ is 

The expression will be $$\cot\left(\frac{\pi}{40}\right)\cot\left(\frac{2\pi}{40}\right)....\cot\left(\frac{19\pi}{40}\right)$$

Now $$\cot A=\tan\left(\frac{\pi}{2}-A\right)$$

Thus $$\cot\frac{\pi}{40}=\tan\left(\frac{\pi}{2}-\frac{\pi}{40}\right)=\tan\frac{19\pi}{40}$$

Also, (cotA)(tanA)=1.

Hence, all the terms other than $$\cot\frac{10\pi}{40}$$ pair up and become 1.

Now $$\cot\frac{10\pi}{40}=\cot\frac{\pi}{4}=1$$

Hence, the value of the expression will be 1.