If A+B=45^{circ} and (cot A-1)(cot B-1)=4 K,t | vedclass

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(C) Given that $A+B=45^{\circ}$.Taking $\cot$ on both sides,we get $\cot(A+B)=\cot 45^{\circ}$.Using the formula $\cot(A+B)=\frac{\cot A \cot B-1}{\co

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$\frac{1}{2}$

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(C) Given that $A+B=45^{\circ}$.Taking $\cot$ on both sides,we get $\cot(A+B)=\cot 45^{\circ}$.Using the formula $\cot(A+B)=\frac{\cot A \cot B-1}{\cot A+\cot B}$,we have:$\frac{\cot A \cot B-1}{\cot A+\cot B}=1$.$\cot A \cot B-1 = \cot A+\cot B$.Rearranging the terms,we get $\cot A \cot B - \cot A - \cot B = 1$.Adding $1$ to both sides to factorize:$\cot A \cot B - \cot A - \cot B + 1 = 1 + 1$.$(\cot A-1)(\cot B-1) = 2$.Given that $(\cot A-1)(\cot B-1)=4K$,we equate the two expressions:$4K = 2$.$K = \frac{2}{4} = \frac{1}{2}$.

If $A+B=45^{\circ}$ and $(\cot A-1)(\cot B-1)=4 K$,then $K=$

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