Question

The maximum value of $\left(\cos {\alpha }_1\right).\left(\cos {\alpha }_2\right).......\left(\cos {\alpha }_n\right)$, under the restriction $0\le {\alpha }_1,{\alpha }_2,......,{\alpha }_n\le \frac{\pi }{2}$ and $\cot {\alpha }_1.\cot {\alpha }_2.......\cot {\alpha }_n=1$ is:

• A. $\frac{1}{2^{\frac{n}{2}}}$
• B. $\frac{1}{2^n}$
• C. $\frac{1}{2n}$
• D. $1$

Answer 1

The correct option is A $\frac{1}{2^{\frac{n}{2}}}$

Given condition $\cot {\alpha }_1.\cot {\alpha }_2.............\cot {\alpha }_n=1$

$\Rightarrow \cos {\alpha }_1.\cos {\alpha }_2.......\cos {\alpha }_n=\sin {\alpha }_1.\sin {\alpha }_2.........\sin {\alpha }_n$

This is possible only when ${\alpha }_1={\alpha }_2={\alpha }_3=....={\alpha }_n={45}^o$

$\Rightarrow \cos {\alpha }_1.\cos {\alpha }_2.......\cos {\alpha }_n=\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{2}}.........\frac{1}{\sqrt{2}}=\frac{1}{2^{\frac{n}{2}}}$