d/(dx)  { tan^-1( secx + tanx) } =?  

প্রশ্ন: \( \frac{d}{dx} \tan^{-1}(\sec x + \tan x) = ? \)

সমাধান:

ধরি, \( y = \tan^{-1}(\sec x + \tan x) \)

আমরা জানি, \( \sec x = \frac{1}{\cos x} \) এবং \( \tan x = \frac{\sin x}{\cos x} \)

সুতরাং, \( y = \tan^{-1}\left(\frac{1}{\cos x} + \frac{\sin x}{\cos x}\right) = \tan^{-1}\left(\frac{1 + \sin x}{\cos x}\right) \)

এখন, \( 1 + \sin x = \sin^2\frac{x}{2} + \cos^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2} = \left(\sin\frac{x}{2} + \cos\frac{x}{2}\right)^2 \)

এবং, \( \cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2} = \left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right) \)

তাহলে, \( y = \tan^{-1}\left(\frac{\left(\sin\frac{x}{2} + \cos\frac{x}{2}\right)^2}{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)}\right) \)

\( y = \tan^{-1}\left(\frac{\sin\frac{x}{2} + \cos\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}\right) \)

হর এবং লবকে \( \cos\frac{x}{2} \) দিয়ে ভাগ করে পাই,

\( y = \tan^{-1}\left(\frac{\tan\frac{x}{2} + 1}{1 - \tan\frac{x}{2}}\right) \)

আমরা জানি, \( \tan\frac{\pi}{4} = 1 \), সুতরাং

\( y = \tan^{-1}\left(\frac{\tan\frac{x}{2} + \tan\frac{\pi}{4}}{1 - \tan\frac{x}{2}\tan\frac{\pi}{4}}\right) \)

\( y = \tan^{-1}\left(\tan\left(\frac{x}{2} + \frac{\pi}{4}\right)\right) \)

\( y = \frac{x}{2} + \frac{\pi}{4} \)

এখন, \( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{x}{2} + \frac{\pi}{4}\right) = \frac{1}{2} + 0 = \frac{1}{2} \)

অতএব, \( \frac{d}{dx} \tan^{-1}(\sec x + \tan x) = \frac{1}{2} \)