Question

The value of $\cot \left(\frac{\mathrm{\pi }}{4}+x\right)\cot \left(\frac{\mathrm{\pi }}{4}-x\right)$ is ___________.

Answer 1

$$\begin{gathered} \cot \left(\frac{\mathrm{\pi }}{4}+x\right)\cot \left(\frac{\mathrm{\pi }}{4}-x\right) \\ \frac{1}{\tan (\mathrm{\pi }/4+\mathrm{x})}\times \frac{1}{\tan \left(\mathrm{\pi }/4-\mathrm{x}\right)} \\ \mathrm{using}\mathrm{identities}: \end{gathered}$$
$\begin{array}{rcl}\tan \left(a+b\right)& =& \frac{\tan a+\tan b}{1-\tan a\tan b}\\ \mathrm{and}\tan \left(a-b\right)& =& \frac{\tan a-\tan b}{1+\tan a\tan b}\\ & =& \frac{1}{{\displaystyle \frac{\mathrm{tan\pi }/4+\tan x}{1-\mathrm{tan\pi }/4\tan x}}}\times \frac{1}{{\displaystyle \frac{\mathrm{tan\pi }/4-\tan x}{1+\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\tan x}}}\\ & =& \frac{1-\tan x}{1+\tan x}\times \frac{1+\tan x}{1-\tan x}\\ & =& 1\end{array}$
$\mathrm{Hence},\mathrm{value}\mathrm{of}\cot \left(\frac{\mathrm{\pi }}{4}+x\right)\cot \left(\frac{\mathrm{\pi }}{4}-x\right)\mathrm{is}1.$