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Given, tan-1 1 + tan-1 2 + tan-1 3
= tan-1 {(1 + 2)/(1 - 1*2)} + tan-1 3
= tan-1 {3/(-1} + tan-1 3
= tan-1 (-3) + tan-1 3
= tan-1 {(-3 + 3)/(1 + 3*3)}
= tan-1 (0)
= π
Hence, tan-1 1 + tan-1 2 + tan-1 3 = π