Question

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

Answer 1

Answer: 0

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