tan(π/28) tan((3π)/28) tan((5π)/28).... .tan((13π)/28)=? 

Explanation:

Another Explanation (5): ```html

প্রশ্ন: \( \tan(\frac{\pi}{28}) \tan(\frac{3\pi}{28}) \tan(\frac{5\pi}{28}) \dots \tan(\frac{13\pi}{28}) = ? \)

সমাধান:

আমরা জানি, \(\tan(\frac{\pi}{2} - x) = \cot(x) = \frac{1}{\tan(x)}\). 🤔

এখানে আমাদের কাছে আছে:

\[ \tan(\frac{\pi}{28}) \tan(\frac{3\pi}{28}) \tan(\frac{5\pi}{28}) \tan(\frac{7\pi}{28}) \tan(\frac{9\pi}{28}) \tan(\frac{11\pi}{28}) \tan(\frac{13\pi}{28}) \]

আমরা লিখতে প???রি:

\[ \frac{\pi}{2} - \frac{13\pi}{28} = \frac{14\pi - 13\pi}{28} = \frac{\pi}{28} \\ \frac{\pi}{2} - \frac{11\pi}{28} = \frac{14\pi - 11\pi}{28} = \frac{3\pi}{28} \\ \frac{\pi}{2} - \frac{9\pi}{28} = \frac{14\pi - 9\pi}{28} = \frac{5\pi}{28} \\ \frac{\pi}{2} - \frac{7\pi}{28} = \frac{14\pi - 7\pi}{28} = \frac{7\pi}{28} \]

সুতরাং, আমরা পাই:

\[ \tan(\frac{13\pi}{28}) = \tan(\frac{\pi}{2} - \frac{\pi}{28}) = \cot(\frac{\pi}{28}) = \frac{1}{\tan(\frac{\pi}{28})} \\ \tan(\frac{11\pi}{28}) = \tan(\frac{\pi}{2} - \frac{3\pi}{28}) = \cot(\frac{3\pi}{28}) = \frac{1}{\tan(\frac{3\pi}{28})} \\ \tan(\frac{9\pi}{28}) = \tan(\frac{\pi}{2} - \frac{5\pi}{28}) = \cot(\frac{5\pi}{28}) = \frac{1}{\tan(\frac{5\pi}{28})} \\ \tan(\frac{7\pi}{28}) = \tan(\frac{\pi}{4}) = 1 \]

তাহলে, আমাদের রাশিমালাটি দাঁড়ায়:

\[ \tan(\frac{\pi}{28}) \tan(\frac{3\pi}{28}) \tan(\frac{5\pi}{28}) \tan(\frac{7\pi}{28}) \tan(\frac{9\pi}{28}) \tan(\frac{11\pi}{28}) \tan(\frac{13\pi}{28}) \\ = \tan(\frac{\pi}{28}) \tan(\frac{3\pi}{28}) \tan(\frac{5\pi}{28}) \cdot 1 \cdot \frac{1}{\tan(\frac{5\pi}{28})} \cdot \frac{1}{\tan(\frac{3\pi}{28})} \cdot \frac{1}{\tan(\frac{\pi}{28})} \\ = 1 \]

অতএব, উত্তর 1. 🎉

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