To proove tan~(1)+tan~(2)+tan~(3)=45°

Question

Inverse circular function

Answer 1

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.