sin^2(cos^-1(1/3))-cos^2(sin^-1(1/sqrt3))=? 

প্রশ্ন: \( \sin^2(\cos^{-1}(\frac{1}{3})) - \cos^2(\sin^{-1}(\frac{1}{\sqrt{3}})) = ? \)

ধরি, \( \cos^{-1}(\frac{1}{3}) = \theta \), তাহলে, \( \cos(\theta) = \frac{1}{3} \)।

আমরা জানি, \( \sin^2(\theta) + \cos^2(\theta) = 1 \)। সুতরাং, \( \sin^2(\theta) = 1 - \cos^2(\theta) \)।

অতএব, \( \sin^2(\theta) = 1 - (\frac{1}{3})^2 = 1 - \frac{1}{9} = \frac{8}{9} \)।

সুতরাং, \( \sin^2(\cos^{-1}(\frac{1}{3})) = \frac{8}{9} \).

আবার ধরি, \( \sin^{-1}(\frac{1}{\sqrt{3}}) = \phi \), তাহলে, \( \sin(\phi) = \frac{1}{\sqrt{3}} \)।

আমরা জানি, \( \sin^2(\phi) + \cos^2(\phi) = 1 \)। সুতরাং, \( \cos^2(\phi) = 1 - \sin^2(\phi) \)।

অতএব, \( \cos^2(\phi) = 1 - (\frac{1}{\sqrt{3}})^2 = 1 - \frac{1}{3} = \frac{2}{3} \)।

সুতরাং, \( \cos^2(\sin^{-1}(\frac{1}{\sqrt{3}})) = \frac{2}{3} \).

এখন, \( \sin^2(\cos^{-1}(\frac{1}{3})) - \cos^2(\sin^{-1}(\frac{1}{\sqrt{3}})) = \frac{8}{9} - \frac{2}{3} = \frac{8}{9} - \frac{6}{9} = \frac{2}{9} \)।

সুতরাং, উত্তর: \( \frac{2}{9} \) 🎉