If 1-5-1-10-1-15-1-40-2k- then the value of k is

(1 θ)(1 (45^∘ θ)) =(1 θ)(1 (θ+1θ 1)) =2 So, (1 5^∘)(1 10^∘)(1 15^∘)...(1 40^∘) =(1 5^∘)(1 40^∘)(1 10^∘)(1 35^∘) (1 15^∘)(1 30^∘)(1 20^∘)(1 25^∘) =2^4 ∴ k ...

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Standard XIII

Mathematics

Trigonometric Ratios of Compound Angles

If 1-5∘1-10∘1...

Question

If $(1 - cot 5^{\circ}) (1 - cot 10^{\circ}) (1 - cot 15^{\circ}) \dots (1 - cot 40^{\circ}) = 2^{k}$ , then the value of $k$ is

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Solution

$(1 - cot θ) (1 - cot (45^{\circ} - θ)) = (1 - cot θ) (1 - (\frac{cot θ + 1}{cot θ - 1})) = 2$

So,
$(1 - cot 5^{\circ}) (1 - cot 10^{\circ}) (1 - cot 15^{\circ}) . . . (1 - cot 40^{\circ})$
$= (1 - cot 5^{\circ}) (1 - cot 40^{\circ}) (1 - cot 10^{\circ}) (1 - cot 35^{\circ}) (1 - cot 15^{\circ}) (1 - cot 30^{\circ}) (1 - cot 20^{\circ}) (1 - cot 25^{\circ}) = 2^{4} ∴ k = 4$

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Trigonometric Ratios of Compound Angles

Standard XIII Mathematics

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If (1−cot5∘)(1−cot10∘)(1−cot15∘)⋯(1−cot40∘)=2k, then the value of k is

(1−cotθ)(1−cot(45∘−θ))=(1−cotθ)(1−(cotθ+1cotθ−1))=2 So, (1−cot5∘)(1−cot10∘)(1−cot15∘)...(1−cot40∘) =(1−cot5∘)(1−cot40∘)(1−cot10∘)(1−cot35∘) (1−cot15∘)(1−cot30∘)(1−cot20∘)(1−cot25∘)=24∴k=4

If $(1 - cot 5^{\circ}) (1 - cot 10^{\circ}) (1 - cot 15^{\circ}) \dots (1 - cot 40^{\circ}) = 2^{k}$ , then the value of $k$ is