মান কোনটি নির্নয় করঃ  d/dx {tan^-1(cosx/(1+sinx))} 

প্রশ্ন: \( \frac{d}{dx} \tan^{-1}\left(\frac{\cos x}{1+\sin x}\right) \) এর মান নির্ণয় করো।

সমাধান:

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

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

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

আমরা জানি, \( \sin 2\theta = 2\sin\theta\cos\theta \) এবং \( 1 - \cos 2\theta = 2\sin^2\theta \)

অতএব, \( \sin\left(\frac{\pi}{2} - x\right) = 2\sin\left(\frac{\pi}{4} - \frac{x}{2}\right)\cos\left(\frac{\pi}{4} - \frac{x}{2}\right) \)
এবং \( 1 - \cos\left(\frac{\pi}{2} - x\right) = 2\sin^2\left(\frac{\pi}{4} - \frac{x}{2}\right) \)

তাহলে, \( y = \tan^{-1}\left(\frac{2\sin\left(\frac{\pi}{4} - \frac{x}{2}\right)\cos\left(\frac{\pi}{4} - \frac{x}{2}\right)}{2\sin^2\left(\frac{\pi}{4} - \frac{x}{2}\right)}\right) \)

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

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

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

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

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

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

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

\( \frac{dy}{dx} = 0 + \frac{1}{2} = \frac{1}{2} \)

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অন্যভাবে:

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

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

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

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

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

\( \frac{dy}{dx} = -\frac{1}{2} \)

সুতরাং, \( \frac{d}{dx} \tan^{-1}\left(\frac{\cos x}{1+\sin x}\right) = -\frac{1}{2} \)

উত্তর: \( -\frac{1}{2} \) 🎉🎉🎉