The correct option is D a2(π2n) ∠A_1OA_2=2πn ∴A_1A_2=√(R^2+R^2 2R.Rcos(2πn)) Since cos2θ=1 2sin^2θ we have cos(2πn)=1 2sin^2πn nbsp ...

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Chain Rule of Differentiation

The value of ...

Question

The value of $r + R$ is

\frac{a}{2} cot (\frac{π}{n})

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a cot (\frac{π}{2 n})

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\frac{a}{4} cot (\frac{π}{2 n})

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\frac{a}{2} cot (\frac{π}{2 n})

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Solution

The correct option is D $\frac{a}{2} cot (\frac{π}{2 n})$
$∠ A_{1} O A_{2} = \frac{2 π}{n}$
$∴ A_{1} A_{2} = \sqrt{R^{2} + R^{2} - 2 R . R cos (\frac{2 π}{n})}$
Since $c o s 2 θ = 1 - 2 {sin}^{2} θ$ we have $cos (\frac{2 π}{n}) = 1 - 2 {sin}^{2} \frac{π}{n}$ $\Rightarrow A_{1} A_{2} = \sqrt{2 R^{2} - 2 R^{2} (1 - 2 {sin}^{2} \frac{π}{n})}$
$= \sqrt{4 R^{2} {sin}^{2} (\frac{π}{n})}$ $= 2 R sin \frac{π}{n}$ ........... $(1)$
Consider $A_{1} A_{3}$
$= \sqrt{R^{2} + R^{2} - 2 R . R cos (\frac{4 π}{n})}$ $= \sqrt{2 R^{2} - 2 R^{2} (1 - 2 {sin}^{2} \frac{4 π}{n})}$
$= \sqrt{4 R^{2} {sin}^{2} (\frac{4 π}{n})}$
$= 2 R sin \frac{2 π}{n}$ ........... $(2)$
Continuing like this $A_{1} A_{4} = 2 R sin (\frac{3 π}{n})$ and so on.
$\frac{π}{n}, \frac{2 π}{n}, \frac{3 π}{n} . . .$ are in A.P
$∴ {(j - 1)}^{t h}$ term is
$= \frac{π}{n} + (j - 2) \frac{π}{n} = (j - 1) \frac{π}{n}$
$∴ A_{i} A_{j} = 2 R sin (j - 1) \frac{π}{n}$ where $j = 1, 2, 3, . . . n$
$∵ ∠ A_{1} I A_{2} = \frac{2 π}{n}$
In $D I A_{1},$
$tan (\frac{π}{n}) = \frac{A_{1} D}{r} = \frac{\frac{a}{2}}{r}$
$∴ r = \frac{a}{2} cot (\frac{π}{n})$ .............. $(3)$
From eqn $(2)$
$A_{1} A_{2} = 2 R sin (\frac{π}{n})$
or $a = 2 R sin (\frac{π}{n})$
$∴ R = \frac{a}{2 sin (\frac{π}{n})}$
From eqn $(3)$
$r = \frac{a cos (\frac{π}{n})}{2 sin (\frac{π}{n})}$
$∴ r + R = \frac{a}{2 sin (\frac{π}{n})} [1 + cos (\frac{π}{n})]$
Using multiple angle formulae, we have
$= \frac{a (2 {cos}^{2} (\frac{π}{2 n}))}{4 sin (\frac{π}{2 n}) cos (\frac{π}{2 n})}$
$∴ r + R = \frac{a}{2} cot (\frac{π}{2 n})$

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