If f(θ)=cotθ1+cotθ and α+β=5π4, then the value of f(α).f(β) is

cot(α+β)=cot5π4=1f(α).f(β)=cotα.cotβ(1+cotα)(1+cotβ)=cotα.cotβ1+(cotα+cotβ)+cotα.cotβ [∵1+cotα+cotβ=cotα.cotβ]=cotα.cotβ2.cotα.cotβ=12

If f(θ)=cotθ1+cotθ and α+β=5π4, then the value of f(α).f(β) is

cot(α+β)=cot5π4=1f(α).f(β)=cotα.cotβ(1+cotα)(1+cotβ)=cotα.cotβ1+(cotα+cotβ)+cotα.cotβ [∵1+cotα+cotβ=cotα.cotβ]=cotα.cotβ2.cotα.cotβ=12

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If $f (θ) = \frac{\cot θ}{1 + \cot θ}$ and $α + β = \frac{5 π}{4}$ , then the value of $f (α) . f (β)$ is

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$- \frac{1}{2}$

$\frac{1}{2}$

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Detailed Solution

$\cot (α + β) = \cot \frac{5 π}{4} = 1$

$f (α) . f (β) = \frac{\cot α . \cot β}{(1 + \cot α) (1 + \cot β)}$

$= \frac{\cot α . \cot β}{1 + (\cot α + \cot β) + \cot α . \cot β} [∵ 1 + \cot α + \cot β = \cot α . \cot β]$

$= \frac{\cot α . \cot β}{2. \cot α . \cot β}$

$= \frac{1}{2}$

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If f(θ)=cotθ1+cotθ and α+β=5π4, then the value of f(α).f(β) is

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If $f (θ) = \frac{\cot θ}{1 + \cot θ}$ and $α + β = \frac{5 π}{4}$ , then the value of $f (α) . f (β)$ is