cot frac{pi}{16} cdot cot frac{2 pi}{16} cdot | vedclass

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(B) Let $P = \cot \frac{\pi}{16} \cdot \cot \frac{2 \pi}{16} \cdot \cot \frac{3 \pi}{16} \cdot \cot \frac{4 \pi}{16} \cdot \cot \frac{5 \pi}{16} \cdot

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(B) Let $P = \cot \frac{\pi}{16} \cdot \cot \frac{2 \pi}{16} \cdot \cot \frac{3 \pi}{16} \cdot \cot \frac{4 \pi}{16} \cdot \cot \frac{5 \pi}{16} \cdot \cot \frac{6 \pi}{16} \cdot \cot \frac{7 \pi}{16}$.Using the property $\cot(\frac{\pi}{2} - 	heta) = 	an 	heta$,we can rewrite the last three terms:$\cot \frac{5 \pi}{16} = \cot(\frac{\pi}{2} - \frac{3 \pi}{16}) = 	an \frac{3 \pi}{16}$$\cot \frac{6 \pi}{16} = \cot(\frac{\pi}{2} - \frac{2 \pi}{16}) = 	an \frac{2 \pi}{16}$$\cot \frac{7 \pi}{16} = \cot(\frac{\pi}{2} - \frac{\pi}{16}) = 	an \frac{\pi}{16}$Substituting these back into the expression:$P = (\cot \frac{\pi}{16} \cdot 	an \frac{\pi}{16}) \cdot (\cot \frac{2 \pi}{16} \cdot 	an \frac{2 \pi}{16}) \cdot (\cot \frac{3 \pi}{16} \cdot 	an \frac{3 \pi}{16}) \cdot \cot \frac{4 \pi}{16}$Since $\cot 	heta \cdot 	an 	heta = 1$:$P = 1 \cdot 1 \cdot 1 \cdot \cot \frac{4 \pi}{16} = \cot \frac{\pi}{4} = 1$.

$\cot \frac{\pi}{16} \cdot \cot \frac{2 \pi}{16} \cdot \cot \frac{3 \pi}{16} \cdot \cot \frac{4 \pi}{16} \cdot \cot \frac{5 \pi}{16} \cdot \cot \frac{6 \pi}{16} \cdot \cot \frac{7 \pi}{16} = $

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