If -1-2-n - 1 and 0- -1- -2- -n - -2- then the maximum value of cos-1 cos-2-cos-n- is

Given: nbsp;(α_1)(α_2)...(α_n) = 1 ⇒ (cosα_1) (cosα_2)...(cosα_n) = (sinα_1)(sinα_2)...(sinα_n) Assuming nbsp; y = (cosα_1) (cosα_2)...(cosα_n) Squaring bot ...

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Standard XI

Mathematics

Conditional Identities

If α1α2...αn ...

Question

If $(cot α_{1}) (cot α_{2}) . . . (cot α_{n}) = 1$ and $0 < α_{1}, α_{2}, . . ., α_{n} < \frac{π}{2}$ , then the maximum value of $(cos α_{1}) (cos α_{2}) . . . (cos α_{n}),$ is

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

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

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\frac{1}{2 n}

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1

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Solution

The correct option is A $\frac{1}{2^{n / 2}}$
Given: $(cot α_{1}) (cot α_{2}) . . . (cot α_{n}) = 1$
$\Rightarrow (cos α_{1}) (cos α_{2}) . . . (cos α_{n}) = (sin α_{1}) (sin α_{2}) . . . (sin α_{n})$
Assuming
$y = (cos α_{1}) (cos α_{2}) . . . (cos α_{n})$
Squaring both sides, we get
$\Rightarrow y^{2} = ({cos}^{2} α_{1}) ({cos}^{2} α_{2}) . . . ({cos}^{2} α_{n}) \Rightarrow y^{2} = cos α_{1} sin α_{1} cos α_{2} sin α_{2} . . . cos α_{n} sin α_{n} \Rightarrow y^{2} = \frac{1}{2^{n}} [sin 2 α_{1} sin 2 α_{2} . . . sin 2 α_{n}]$

As $0 < α_{1}, α_{2}, . . ., < \frac{π}{2}$
$\Rightarrow 0 < 2 α_{1}, 2 α_{2}, . . .2 α_{2} < π$
So,
$0 < sin 2 α_{1} sin 2 α_{2} . . . sin 2 α_{n} \leq 1 \Rightarrow 0 < y^{2} \leq \frac{1}{2^{n}}$

Therefore, the maximum value of $y$ is $\frac{1}{2^{n / 2}}$

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If (cotα1)(cotα2)...(cotαn)=1 and 0<α1,α2,...,αn<π2, then the maximum value of (cosα1)(cosα2)...(cosαn), is

If $(cot α_{1}) (cot α_{2}) . . . (cot α_{n}) = 1$ and $0 < α_{1}, α_{2}, . . ., α_{n} < \frac{π}{2}$ , then the maximum value of $(cos α_{1}) (cos α_{2}) . . . (cos α_{n}),$ is