Question

Given $f(x)=\tan ^{-1}(\cot x)+\cot ^{-1}(\tan x),\left(\frac{\pi}{2}< x < \pi\right)$, then $\left|f'\left(\frac{2 \pi}{3}\right)-f'\left(\frac{5 \pi}{6}\right)\right|$ is equal to

Answer 1

$f(x)=\pi-\left(\cot ^{-1}(\cot x)+\tan ^{-1}(\tan x)\right)$
$=\pi-(x+x-\pi)=2 \pi-2 x$
$\Rightarrow f^{\prime}(x)=-2=$ constant
$\Rightarrow\left|f^{\prime}\left(\frac{2 \pi}{3}\right)-f^{\prime}\left(\frac{5 \pi}{6}\right)\right|=0$