Question
The maximum value of $$\left( {\cot \,{\alpha _1}} \right).\left( {\cot \,{\alpha _2}} \right)\,.....\,\left( {\cot \,{\alpha _n}} \right),$$ ; under the restrictions $$0 \leqslant {\alpha _1},{\alpha _2},.....,{\alpha _n} \leqslant \frac{\pi }{2}\,{\text{and}}$$ $$\left( {\cot \,{\alpha _1}} \right).\left( {\cot \,{\alpha _2}} \right)\,.....\,\left( {\cot \,{\alpha _n}} \right) = 1$$ is
A.
$$\frac{1}{{{2^{\frac{n}{2}}}}}$$
B.
$$\frac{1}{{{2^n}}}$$
C.
$$\frac{1}{{2n}}$$
D.
1
Answer :
$$\frac{1}{{{2^{\frac{n}{2}}}}}$$
Solution :
We are given that
$$\eqalign{
& \left( {\cot \,{\alpha _1}} \right).\left( {\cot \,{\alpha _2}} \right)\,.....\,\left( {\cot \,{\alpha _n}} \right) = 1 \cr
& \Rightarrow \,\left( {\cos \,{\alpha _1}} \right)\left( {\cos \,{\alpha _2}} \right)\,.....\,\left( {\cos \,{\alpha _n}} \right) \cr
& = \left( {\sin \,{\alpha _1}} \right)\,\left( {\sin \,{\alpha _2}} \right)\,.....\left( {\sin \,{\alpha _n}} \right)\,\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr
& {\text{Let}}\,y\,{\text{ = }}\left( {\cos \,{\alpha _1}} \right)\,\left( {\cos \,{\alpha _2}} \right)....\,\left( {\cos \,{\alpha _n}} \right)\,\left( {{\text{to}}\,{\text{be}}\,{\text{max}}{\text{.}}} \right) \cr
& {\text{Squaring}}\,{\text{both}}\,{\text{sides,}}\,{\text{we}}\,{\text{get}} \cr
& {y^{\text{2}}}{\text{ = }}\,\left( {{{\cos }^2}\,{\alpha _1}} \right)\,\,\left( {{{\cos }^2}\,{\alpha _2}} \right)\,.....\,\left( {{{\cos }^2}\,{\alpha _n}} \right) \cr
& = \,\cos \,{\alpha _1}\,\sin \,{\alpha _1}\,\cos \,{\alpha _2}\,\sin \,{\alpha _2}\,.....\,\cos \,{\alpha _n}\,\sin \,{\alpha _n}\,\,\,\,\left( {{\text{Using}}\,\left( 1 \right)} \right) \cr
& = \,\frac{1}{{{2^n}}}\,\left[ {\sin \,2{\alpha _1}\,\sin \,2{\alpha _2}\,.....\sin 2\,{\alpha _n}} \right] \cr
& {\text{As}}\,\,0 \leqslant {\alpha _1},\,{\alpha _2},.....{\alpha _n} \leqslant \frac{\pi }{2} \cr
& \therefore \,\,\,0 \leqslant 2{\alpha _1},\,2{\alpha _2},.....\sin \,2{\alpha _n}\, \leqslant \pi \cr
& \Rightarrow \,\,0 \leqslant \,\sin \,2{\alpha _1},\,\sin \,2{\alpha _2},.....\,\sin \,2{\alpha _n}\, \leqslant 1 \cr
& \therefore \,{y^2}\, \leqslant \,\frac{1}{{{2^n}}}.1 \cr
& \Rightarrow \,\,y \leqslant \,\frac{1}{{{2^{\frac{n}{2}}}}} \cr} $$
∴ Max. value of $$y$$ is $$\frac{1}{{{2^{\frac{n}{2}}}}}.$$