\( \tan(\cos^{-1} \sin \cot^{-1} (\frac{3}{4})) = ? \)

প্রশ্ন: \( \tan(\cos^{-1} \sin \cot^{-1} (\frac{3}{4})) = ? \)

উত্তর: \( \frac{3}{4} \)

সমাধান:

ধরি, \( \theta = \cot^{-1} \left( \frac{3}{4} \right) \)

তাহলে, \( \cot \theta = \frac{3}{4} \)

এখন, \(\cot \theta = \frac{\cos \theta}{\sin \theta}\), সুতরাং:

\( \frac{\cos \theta}{\sin \theta} = \frac{3}{4} \)

অর্থাৎ, \(\cos \theta = 3k\) এবং \(\sin \theta = 4k\) জন্য কিছু ধনাত্মক \(k\)।

প্রতিসূত্রে, \(\sin^2 \theta + \cos^2 \theta = 1\), সুতরাং:

\( (4k)^2 + (3k)^2 = 1 \)

\( 16k^2 + 9k^2 = 1 \)

\( 25k^2 = 1 \)

অতএব, \( k = \pm \frac{1}{5} \)

ধরা হয়, \(\theta\) প্রথম কোণে, তাহলে \(k = \frac{1}{5}\)।

অর্থাৎ:

\(\cos \theta = 3 \times \frac{1}{5} = \frac{3}{5}\)

\(\sin \theta = 4 \times \frac{1}{5} = \frac{4}{5}\)

এখন, \(\sin \cot^{-1} \left( \frac{3}{4} \right) = \sin \theta = \frac{4}{5}\).

অতএব, আমাদের মূল অভিব্যক্তি এখন:

\( \tan \left( \cos^{-1} \left( \frac{4}{5} \right) \right) \)

ধরি, \( \phi = \cos^{-1} \left( \frac{4}{5} \right) \), তাহলে:

\(\cos \phi = \frac{4}{5}\)

অর্থাৎ, \(\sin \phi = \sqrt{1 - \left( \frac{4}{5} \right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}\)

সুতরাং, \(\tan \phi = \frac{\sin \phi}{\cos \phi} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}\)

উত্তর:

\( \boxed{\frac{3}{4}} \)