Question

The maximum value of $\cos {\alpha }_1.\cos {\alpha }_2......\cos {\alpha }_n\forall 0\le {\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n\le \frac{\pi }{2}\&\cot {\alpha }_1.\cot {\alpha }_2......\cot {\alpha }_n=1$
is:

Answer 1

Given $\left(\cot {\alpha }_1\right).\left(\cot {\alpha }_2\right).....\left(\cot {\alpha }_n\right)=1$
$\therefore \cos {\alpha }_1.\cos {\alpha }_2...\cos {\alpha }_n=\sin {\alpha }_1.\sin {\alpha }_2...\sin {\alpha }_n$
Now,
$(\cos {\alpha }_1.\cos {\alpha }_2...\cos {\alpha }_n{)}^2$
$=(\cos {\alpha }_1.\cos {\alpha }_2...\cos {\alpha }_n).(\cos {\alpha }_1.\cos {\alpha }_2...\cos {\alpha }_n)$
$=(\cos {\alpha }_1.\cos {\alpha }_2...\cos {\alpha }_n).(\sin {\alpha }_1.\sin {\alpha }_2...\sin {\alpha }_n)$
$=\frac{1}{2^n}\sin 2{\alpha }_1.\sin 2{\alpha }_2...\sin 2{\alpha }_n.$

$\because 0\le {\alpha }_1,{\alpha }_2,.....,{\alpha }_n\le \frac{\pi }{2}\therefore 0\le 2{\alpha }_1,2{\alpha }_2,.....,2{\alpha }_n\le \pi$
So each of $\sin 2{\alpha }_1\le 1$
$\therefore (\cos {\alpha }_1.\cos {\alpha }_2...\cos {\alpha }_n{)}^2\le \frac{1}{2^n}.$
But each of $\cos {\alpha }_1$ is positive.

$\therefore \cos {\alpha }_1.\cos {\alpha }_2....\cos {\alpha }_n\le \sqrt{\frac{1}{2^n}}\Rightarrow \cos {\alpha }_1.\cos {\alpha }_2....\cos {\alpha }_n\le \frac{1}{2^{\frac{n}{2}}}.$
So the maximum value of
$\cos {\alpha }_1.\cos {\alpha }_2......\cos {\alpha }_n=\frac{1}{2^{\frac{n}{2}}}$

Hence the correct answer is Option A.