A) \[\frac{1}{{{2}^{n/2}}}\]
B) \[\frac{1}{{{2}^{n}}}\]
C) \[\frac{1}{2n}\]
D) \[1\]
Correct Answer: A
Solution :
We are given that, \[(\cot {{\alpha }_{1}})\cdot (\cot {{\alpha }_{2}})...(\cot {{\alpha }_{n}})=1\] \[\Rightarrow \]\[(\cos {{\alpha }_{1}}\cdot (\cos {{\alpha }_{2}})...(\cos {{\alpha }_{n}})=(\sin {{\alpha }_{1}})\] \[\cdot (\sin {{\alpha }_{2}})...(\sin {{\alpha }_{n}})\] ? (i) Let\[y=(\cos {{\alpha }_{1}})\cdot (\cos {{\alpha }_{2}})...(\cos {{\alpha }_{n}})\] (to be max) Squaring both sides, we get \[{{y}^{2}}=({{\cos }^{2}}{{\alpha }_{1}})\cdot ({{\cos }^{2}}{{\alpha }_{2}})...({{\cos }^{2}}{{\alpha }_{n}})\] \[=\cos {{\alpha }_{1}}\cdot \sin {{\alpha }_{1}}\cdot \cos {{\alpha }_{2}}\cdot \sin {{\alpha }_{2}}...\cos {{\alpha }_{n}}\cdot \sin {{\alpha }_{n}}\] [using (i)] \[=\frac{1}{{{2}^{n}}}[\sin 2{{\alpha }_{1}}\cdot \sin 2{{\alpha }_{2}}...\sin 2{{\alpha }_{n}}]\] As \[0\le {{\alpha }_{1}},\,\,{{\alpha }_{2}},...{{\alpha }_{n}}\le \frac{\pi }{2}\] \[\therefore \] \[0\le 2{{\alpha }_{1}},\,\,2{{\alpha }_{2}},...,2{{\alpha }_{n}}\le \pi \] \[\Rightarrow \]\[0\le \sin 2{{\alpha }_{1}},\,\,sin2{{\alpha }_{2}},...\sin 2{{\alpha }_{n}}\le 1\] \[\therefore \] \[{{y}^{2}}\le \frac{1}{{{2}^{n}}}\cdot 1\] \[\Rightarrow y\le \frac{1}{{{2}^{n/2}}}\]
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