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

Another Explanation (5): ```html

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

সমাধান:

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

আমরা জানি, \( \sin^2(\theta) + \cos^2(\theta) = 1 \).

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

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

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

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

সুতরাং, \( \sin^2(\cos^{-1} (\frac{1}{3} )) - \cos^2(\sin^{-1}(\frac{1}{\sqrt{3}})) = \frac{8}{9} - \frac{2}{3} = \frac{8}{9} - \frac{6}{9} = \frac{2}{9} \) 🎉.

অতএব, উত্তর: \( \frac{2}{9} \).

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