If in a triangle ABC,angle C is 45^o,then (1 | vedclass

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(B) Given,$\angle C = 45^o$. Since $A + B + C = 180^o$,we have $A + B = 180^o - 45^o = 135^o$. Using the formula $\cot(A + B) = \frac{\cot A \cot B -

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$2$

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(B) Given,$\angle C = 45^o$. Since $A + B + C = 180^o$,we have $A + B = 180^o - 45^o = 135^o$. Using the formula $\cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}$,we get: $\cot(135^o) = \frac{\cot A \cot B - 1}{\cot A + \cot B}$. Since $\cot(135^o) = -1$,we have: $-1 = \frac{\cot A \cot B - 1}{\cot A + \cot B}$. $-(\cot A + \cot B) = \cot A \cot B - 1$. $1 = \cot A \cot B + \cot A + \cot B$. Adding $1$ to both sides: $1 + 1 = 1 + \cot A + \cot B + \cot A \cot B$. $2 = (1 + \cot A)(1 + \cot B)$. Thus,$(1 + \cot A)(1 + \cot B) = 2$.

If in a triangle $ABC$,angle $C$ is $45^o$,then $(1 + \cot A)(1 + \cot B) = $

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