d/dx[tan^-1(cotx)+cot^-2(tanx)]=?

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

সমাধান:

আমরা জানি, \(\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}\). 🤓

তাহলে, \(\tan^{-1}(\cot x) = \frac{\pi}{2} - \cot^{-1}(\cot x) = \frac{\pi}{2} - x\), যখন \(0 < x < \pi\). 🎉

আবার, \(\cot^{-1}(\tan x) = \frac{\pi}{2} - \tan^{-1}(\tan x) = \frac{\pi}{2} - x\), যখন \(-\frac{\pi}{2} < x < \frac{\pi}{2}\). 🥳

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

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

অতএব, উত্তর: -2। ✅