Question

The maximum value of $\left(\cos \alpha_1\right) \cdot\left(\cos \alpha_2\right) \ldots \ldots \ldots\left(\cos \alpha_n\right)$, under the restrictions $0 \leq \alpha_1, \alpha_2, \ldots \ldots \ldots \ldots, \alpha_n \leq \frac{\pi}{2}$ and $\cot \alpha_1 \cdot \cot \alpha_2 \ldots \ldots \ldots \cot \alpha_n=1$ is

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

Answer 1

Answer: (A)

$\text { Given } \cos \alpha_1 \cos \alpha_2-\cos \alpha_3 \ldots . \cos \alpha_n=\sin \alpha_1 \sin \alpha_2 \ldots \ldots \sin \alpha_n $
$y=\left(\cos \alpha_1\right)\left(\cos \alpha_2\right) \ldots \ldots\left(\cos \alpha_n\right)\left(\sin \alpha_1\right)\left(\sin \alpha_2\right)\left(\sin \alpha_3\right) \ldots\left(\sin \alpha_n\right) $
${[M-1] y^2=\left(\cos \alpha_1\right)\left(\cos \alpha_2\right) \ldots\left(\cos \alpha_n\right)\left(\sin \alpha_1\right)\left(\sin \alpha_2\right) \ldots\left(\sin \alpha_n\right)} $
$ y^2=\frac{1}{2^n}\left(\sin 2 \alpha_1\right)\left(\sin 2 \alpha_1\right) \ldots\left(\sin +\alpha_n\right) $
$\therefore y_{\max }^2=\frac{1}{2^n} \Rightarrow y_{\max }=\frac{1}{2^{4 / 2}}$
${\left[\text { M-2] } \frac{1+\tan ^2 \alpha_1}{2} \geq\left(1+\tan ^2 \alpha\right)^{1 / 2}=\tan \alpha_1\right.} $
$\therefore 1+\tan ^2 \alpha_1 \geq 2 \tan \alpha_1 \sec \alpha_1 \sin \alpha_2 \ldots . \sin \alpha_n \geq 2^{4 / 2} $
$ \sec ^2 \alpha_1 \geq 2 \tan \alpha_1 $
$ \sec ^2 \alpha^2 \geq 2 \tan \alpha_2$
$ \sec ^2 \alpha_n \geq 2 \tan \alpha_n $
$\therefore \cos _1 \cos \alpha_2 \ldots \ldots \cos \alpha_n \leq \frac{1}{2^{n / 2}}$