Cotangent

$(1-cot20^{o})(1-cot25^{o})=x$

$1-(cot20^{o}+cot25^{o})+cot20^{o}cot25^{o}$

using $cot(A+B)=\frac{cotAcotB-1}{cotA+cotB}$

substitute value of $cotA+cotB$

$\large{1-(\frac{cot20^{o}cot25^{o}-1}{cot(20^{o}+25^{o})})+cot20^{o}cot25^{o}}$

As $cot45^{o}=1$ , open the brackets

answer is $\boxed{2}$

$(1-\cot{20})(1-\cot{(45-20)})=(1-\cot{20})\left(1-\frac{(\cot{20})+1}{(\cot{20})-1}\right)$ $=(1-\cot{20})\left(\frac{-2}{(\cot{20})-1}\right)=2$ .

We have used $\cot{(a-b)}=\frac{(\cot{a})(cot{b})+1}{\cot{b}-\cot{a}}$