d/dx[tan^-1(cotx)+cot^-1(tanx)]=? 

প্রশ্ন:

\(\frac{d}{dx}[\tan^{-1}(\cot x) + \cot^{-1}(\tan x)] = ?\)

সমাধান:

আমরা জানি, \(\cot x = \tan\left(\frac{\pi}{2} - x\right)\) এবং \(\tan x = \cot\left(\frac{\pi}{2} - x\right)\)

সুতরাং, \(\tan^{-1}(\cot x) = \tan^{-1}\left(\tan\left(\frac{\pi}{2} - x\right)\right)\)

এবং, \(\cot^{-1}(\tan x) = \cot^{-1}\left(\cot\left(\frac{\pi}{2} - x\right)\right)\)

যদি \(0 < x < \pi\), তবে \(\frac{\pi}{2} - x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), সুতরাং, \(\tan^{-1}\left(\tan\left(\frac{\pi}{2} - x\right)\right) = \frac{\pi}{2} - x\)

আবার, \(0 < x < \pi\) হলে, \(\frac{\pi}{2} - x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), সুতরাং, \(\cot^{-1}\left(\cot\left(\frac{\pi}{2} - x\right)\right) = \frac{\pi}{2} - x\)

অতএব, \(\tan^{-1}(\cot x) + \cot^{-1}(\tan x) = \frac{\pi}{2} - x + \frac{\pi}{2} - x = \pi - 2x\)

এখন, \(\frac{d}{dx}[\tan^{-1}(\cot x) + \cot^{-1}(\tan x)] = \frac{d}{dx}[\pi - 2x] = -2\)

উত্তর:

-2