The value of cot

$\cot \left(\frac{\pi}{4}+x\right) \cot \left(\frac{\pi}{4}-x\right)$

$\frac{1}{\tan (\pi / 4+x)} \times \frac{1}{\tan (\pi / 4-x)}$

using identities :

$\tan (a+b)=\frac{\tan a+\tan b}{1-\tan a \tan b}$

and $\tan (a-b)=\frac{\tan a-\tan b}{1+\tan a \tan b}$

$=\frac{1}{\frac{\tan \pi / 4+\tan x}{1-\tan \pi / 4 \tan x}} \times \frac{1}{\frac{\tan \pi / 4-\tan x}{1+\tan \frac{\pi}{4} \tan x}}$

$=\frac{1-\tan x}{1+\tan x} \times \frac{1+\tan x}{1-\tan x}$

$=1$

Hence, value of $\cot \left(\frac{\pi}{4}+x\right) \cot \left(\frac{\pi}{4}-x\right)$ is 1 .