Dear Student, Question correction:Evaluate : tan-1 (1) + tan-1 (2) + tan-1 (3) - πSolution:We know, tan-1 (x) + tan-1 (y) = tan-1 x+y1-xySo,tan-1 (1) + tan-1 (2) + tan-1 (3) - π= tan-1 1+21-(1×2) + tan-1 (3) - π= tan-1 31-2 + tan-1 (3) - π= tan-1 3-1 + tan-1 (3) - π= tan-1 (-3) + tan-1 (3) - π= tan-1 -3+31-(-3×3)- π= tan-1 01-(-9)- π= tan-1 (0) - π= π - π {Since, tan π = 0}= 0 Regards

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Dear Student,

Q u e s t i o n c o r r e c t i o n : E v a l u a t e : \tan^{- 1} (1) + \tan^{- 1} (2) + \tan^{- 1} (3) - π Solution : We know, \tan^{- 1} (x) + \tan^{- 1} (y) = \tan^{- 1} (\frac{x + y}{1 - xy}) So, \tan^{- 1} (1) + \tan^{- 1} (2) + \tan^{- 1} (3) - π = \tan^{- 1} [\frac{1 + 2}{1 - (1 \times 2)}] + \tan^{- 1} (3) - π = \tan^{- 1} [\frac{3}{1 - 2}] + \tan^{- 1} (3) - π = \tan^{- 1} [\frac{3}{- 1}] + \tan^{- 1} (3) - π = \tan^{- 1} (- 3) + \tan^{- 1} (3) - π = \tan^{- 1} [\frac{- 3 + 3}{1 - (- 3 \times 3)}] - π = \tan^{- 1} [\frac{0}{1 - (- 9)}] - π = \tan^{- 1} (0) - π = π - π {Since, \tan π = 0} = 0

Regards

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