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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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(1 - cot 5^{\circ}) (1 - cot 10^{\circ}) (1 - cot 15^{\circ}) \dots (1 - cot 40^{\circ}) = 2^{k}

, then the value of

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\frac{1}{1! 10!} + \frac{1}{2! 9!} + \frac{1}{3! 10!} + . . . . . + \frac{1}{1! 10!} = \frac{2}{k!} (2^{k - 1} - 1)

then find the value of k.

\frac{1}{1! 10!} + \frac{1}{2! 9!} + \frac{1}{3! 8!} + . . . . . + \frac{1}{1! 10!} = \frac{2}{k!} (2^{k - 1} - 1)

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Trigonometric Ratios of Compound Angles

Standard XII Mathematics

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