If (cot alpha_1)(cot alpha_2) ldots (cot alph | vedclass

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Question

(A) Let $P = \cos \alpha_1 \cos \alpha_2 \ldots \cos \alpha_n$. Then $\frac{1}{P^2} = \frac{1}{\cos^2 \alpha_1 \cos^2 \alpha_2 \ldots \cos^2 \alpha_n}

Vedclass · Accepted Answer

$\frac{1}{2^{n/2}}$

Vedclass · Answer

(A) Let $P = \cos \alpha_1 \cos \alpha_2 \ldots \cos \alpha_n$. Then $\frac{1}{P^2} = \frac{1}{\cos^2 \alpha_1 \cos^2 \alpha_2 \ldots \cos^2 \alpha_n} = \sec^2 \alpha_1 \sec^2 \alpha_2 \ldots \sec^2 \alpha_n$. Using the identity $\sec^2 \alpha = 1 + 	an^2 \alpha$, we have: $\frac{1}{P^2} = (1 + 	an^2 \alpha_1)(1 + 	an^2 \alpha_2) \ldots (1 + 	an^2 \alpha_n)$. By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality, $1 + 	an^2 \alpha_i \geq 2 	an \alpha_i$. Therefore, $\frac{1}{P^2} \geq (2 	an \alpha_1)(2 	an \alpha_2) \ldots (2 	an \alpha_n) = 2^n (	an \alpha_1 	an \alpha_2 \ldots 	an \alpha_n)$. Since $(\cot \alpha_1)(\cot \alpha_2) \ldots (\cot \alpha_n) = 1$, it follows that $(	an \alpha_1)(	an \alpha_2) \ldots (	an \alpha_n) = 1$. Thus, $\frac{1}{P^2} \geq 2^n$, which implies $P^2 \leq \frac{1}{2^n}$. Taking the square root, $P \leq \frac{1}{2^{n/2}}$. The maximum value is $\frac{1}{2^{n/2}}$.

List-$I$	List-$II$
$(I) \ 3 \sin^2 x + 4 \cos^2 x - 2$	$(a) \ [\frac{1}{4}, 1]$
$(II) \ \cos^2 x + \sin^4 x$	$(b) \ [-\frac{1}{4}, \frac{1}{4}]$
$(III) \ \sin^6 x + \cos^6 x$	$(c) \ [1, 2]$
$(IV) \ \cos x \cos(\frac{2 \pi}{3} + x) \cos(\frac{2 \pi}{3} - x)$	$(d) \ [\frac{3}{4}, 1]$
	$(e) \ [0, 1]$

If $(\cot \alpha_1)(\cot \alpha_2) \ldots (\cot \alpha_n) = 1$ where $0 < \alpha_1, \alpha_2, \ldots, \alpha_n < \pi/2$, then the maximum value of $(\cos \alpha_1)(\cos \alpha_2) \ldots (\cos \alpha_n)$ is given by

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