sincot^-1tancos^-1x=?

প্রশ্ন: \( \sin(\cot^{-1}(\tan(\cos^{-1} x))) = ? \) 🧐

সমাধান:

ধরি, \( \cos^{-1} x = \theta \). তাহলে, \( \cos \theta = x \).

অতএব, \( \tan \theta = \frac{\sqrt{1 - \cos^2 \theta}}{\cos \theta} = \frac{\sqrt{1 - x^2}}{x} \) 😊।

সুতরাং, \( \tan(\cos^{-1} x) = \tan \theta = \frac{\sqrt{1 - x^2}}{x} \) 🙌।

এখন, \( \cot^{-1}(\tan(\cos^{-1} x)) = \cot^{-1} \left( \frac{\sqrt{1 - x^2}}{x} \right) \).

ধরি, \( \cot^{-1} \left( \frac{\sqrt{1 - x^2}}{x} \right) = \alpha \). তাহলে, \( \cot \alpha = \frac{\sqrt{1 - x^2}}{x} \).

অতএব, \( \tan \alpha = \frac{x}{\sqrt{1 - x^2}} \).

এখন, \( \sin \alpha = \frac{\tan \alpha}{\sqrt{1 + \tan^2 \alpha}} = \frac{\frac{x}{\sqrt{1 - x^2}}}{\sqrt{1 + \frac{x^2}{1 - x^2}}} \).

\( \sin \alpha = \frac{\frac{x}{\sqrt{1 - x^2}}}{\sqrt{\frac{1 - x^2 + x^2}{1 - x^2}}} = \frac{\frac{x}{\sqrt{1 - x^2}}}{\sqrt{\frac{1}{1 - x^2}}} = \frac{\frac{x}{\sqrt{1 - x^2}}}{\frac{1}{\sqrt{1 - x^2}}} = x \) 🤩।

সুতরাং, \( \sin(\cot^{-1}(\tan(\cos^{-1} x))) = \sin \alpha = x \) 🎉।

অতএব, উত্তর: \( x \) ✅।