If f(x) = frac{cot x}{1 + cot x} and alpha + | vedclass

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(D) Given,$f(x) = \frac{\cot x}{1 + \cot x}$ and $\alpha + \beta = \frac{5 \pi}{4}$.Since $\cot(\alpha + \beta) = \cot(\frac{5 \pi}{4}) = 1$,we have $

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$\frac{1}{2}$

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(D) Given,$f(x) = \frac{\cot x}{1 + \cot x}$ and $\alpha + \beta = \frac{5 \pi}{4}$.Since $\cot(\alpha + \beta) = \cot(\frac{5 \pi}{4}) = 1$,we have $\frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta} = 1$.This implies $\cot \alpha \cot \beta - 1 = \cot \alpha + \cot \beta$,or $\cot \alpha \cot \beta = 1 + \cot \alpha + \cot \beta$.Now,$f(\alpha) f(\beta) = \frac{\cot \alpha}{1 + \cot \alpha} 	imes \frac{\cot \beta}{1 + \cot \beta} = \frac{\cot \alpha \cot \beta}{1 + \cot \alpha + \cot \beta + \cot \alpha \cot \beta}$.Substituting $\cot \alpha + \cot \beta = \cot \alpha \cot \beta - 1$,we get $f(\alpha) f(\beta) = \frac{\cot \alpha \cot \beta}{1 + (\cot \alpha \cot \beta - 1) + \cot \alpha \cot \beta} = \frac{\cot \alpha \cot \beta}{2 \cot \alpha \cot \beta} = \frac{1}{2}$.

If $f(x) = \frac{\cot x}{1 + \cot x}$ and $\alpha + \beta = \frac{5 \pi}{4}$,then the value of $f(\alpha) f(\beta)$ is:

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