Hello Aspirant,

I hope that you are doing well.

Let  tan−1(1)=x⇒1=tanx
Let tan−1(2)=y⇒2=tany
Let tan−1(3)=z⇒3=tanz
tan(x+y+z)=1−tanx.tany−tanx.tanz−tany.tanztanx+tany+tanz−tanx.tany.tanz
=1−1×2−1×3−2×31+2+3−1×2×3
=0
⇒x+y+z=π
tan−1(1)+tan−1(2)+tan−1(3)=π

(  x+y+z  cannot be equal to zero, because tan−1(1)+tan−1(2)+tan−1(3) will have some value greater than zero )

I hope this helps.

Good luck!

Thank you