If A + B = 225^circ , then frac{{cot A}}{{1 + | vedclass

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(D) Let $X = \frac{\cot A}{1 + \cot A} \cdot \frac{\cot B}{1 + \cot B}$.Converting to tangent form: $X = \frac{1/	an A}{1 + 1/	an A} \cdot \frac{1/\

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$1/2$

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(D) Let $X = \frac{\cot A}{1 + \cot A} \cdot \frac{\cot B}{1 + \cot B}$.Converting to tangent form: $X = \frac{1/	an A}{1 + 1/	an A} \cdot \frac{1/	an B}{1 + 1/	an B} = \frac{1}{	an A + 1} \cdot \frac{1}{	an B + 1} = \frac{1}{1 + 	an A + 	an B + 	an A 	an B}$.Given $A + B = 225^\circ$,we have $	an(A + B) = 	an(225^\circ) = 1$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B} = 1$,we get $	an A + 	an B = 1 - 	an A 	an B$.Substituting this into the expression for $X$:$X = \frac{1}{1 + (1 - 	an A 	an B) + 	an A 	an B} = \frac{1}{1 + 1} = \frac{1}{2}$.

If $A + B = 225^\circ ,$ then $\frac{{\cot A}}{{1 + \cot A}} \cdot \frac{{\cot B}}{{1 + \cot B}} = $

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