If this is meant to read tan-1(1)+tan-1(2)+tan-1(3)=45°,
then it is false because tan-1(1)=45°, and tan-1(2)≠-tan-1(3).
In fact, tan-1(1)+tan-1(2)+tan-1(3)=180°. This is why:
tan135°=-1, so if tan(tan-1(2)+tan-1(3))=-1, we have proved that tan-1(2)+tan-1(3)=135°.
Let tanA=2 and tanB=3, so A+B has to be 135°,
tan(A+B)=(tanA+tanB)/(1-tanAtanB)=(2+3)/(1-6)=-1=tan135°. This proves that tan-1(2)+tan-1(3)=135°.
Therefore A+B+C=tan-1(1)+tan-1(2)+tan-1(3)=45+135=180° (=4×45°).
We can however write ¼(tan-1(1)+tan-1(2)+tan-1(3))=45°, which we have now proved.