${tan}^{- 1} \frac{a - b}{1 + a b} + {tan}^{- 1} \frac{b - c}{1 + b c} =$

{tan}^{- 1} a - {tan}^{- 1} b

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{tan}^{- 1} a - {tan}^{- 1} c

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{tan}^{- 1} b - {tan}^{- 1} c

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{tan}^{- 1} c - {tan}^{- 1} a

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Solution

The correct option is C ${tan}^{- 1} a - {tan}^{- 1} c$
The value of ${tan}^{- 1} \frac{a - b}{1 + a b} + {tan}^{- 1} \frac{b - c}{1 + b c} = = {tan}^{- 1} a - {tan}^{- 1} b + {tan}^{- 1} b - {tan}^{- 1} c = {tan}^{- 1} a - {tan}^{- 1} c$

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Similar questions

Evaluate tan^-1(a - b)/(1+ab)+ tan^-1(b - c)/(1+bc)+ tan^-1(c - a)/(1+cb) ; ab,bc,ac>-1

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a, b, c > 0

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