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

Observe the following statements:
$(I)$ In $\triangle ABC$,$b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} = s$
$(II)$ In $\triangle ABC$,$\cot \frac{A}{2} = \frac{b+c}{a} \implies B = 90^{\circ}$
Which of the following is correct?

In $\triangle ABC$,if $B=90^{\circ}$,then $2(r+R)=$

In a $\Delta ABC$,if $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$ and the side $a = 2$,then the area of the triangle is:

In a triangle $ABC$,if $A, B, C$ are in arithmetic progression and $\cos A + \cos B + \cos C = \frac{1 + \sqrt{2} + \sqrt{3}}{2 \sqrt{2}}$,then $\tan A =$

Assertion $(A)$: If $A=15^{\circ}, B=17^{\circ}$ and $C=13^{\circ}$,then $\cot 2A + \cot 2B + \cot 2C = \cot 2A \cot 2B \cot 2C$.
Reason $(R)$: In a $\triangle PQR$,$\tan \frac{P}{2} \tan \frac{Q}{2} + \tan \frac{Q}{2} \tan \frac{R}{2} + \tan \frac{P}{2} \tan \frac{R}{2} = 1$.
The correct option among the following is: