1.

\tan ^{-1}(1)+\tan ^{-1}(2)+\tan ^{-1}(3)=\pi

Answer»

Use the following formula:

tan-1(x) + tan-1(y) = tan-1[(x+y)/(1-xy)]

so

tan-1(1) + tan-1(2) + tan-1(3)

= tan-1(1) + tan-1[(2+3)/1-(2*3)]

=tan-1(1) + tan-1(-1)

apply same formula again,

you will get

tan-1(0) = π

hope it helped you

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