Tan⁡Tan−1⁡1a+b+Tan−1⁡ba2+ab+1=

Tan⁡Tan−1⁡1a+b+Tan−1⁡ba2+ab+1=Tan⁡Tan−1⁡1a+b+Tan−1⁡(a+b)−a1+(a+b)a=Tan⁡cot−1⁡a+b+Tan−1⁡(a+b)−Tan1⁡a =Tanπ2-Tan-1a=Tancot-1a =TanTan-11a=1a

Tan⁡Tan−1⁡1a+b+Tan−1⁡ba2+ab+1=

Tan⁡Tan−1⁡1a+b+Tan−1⁡ba2+ab+1=Tan⁡Tan−1⁡1a+b+Tan−1⁡(a+b)−a1+(a+b)a=Tan⁡cot−1⁡a+b+Tan−1⁡(a+b)−Tan1⁡a =Tanπ2-Tan-1a=Tancot-1a =TanTan-11a=1a

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$Tan [{Tan}^{- 1} (\frac{1}{a + b}) + {Tan}^{- 1} (\frac{b}{a^{2} + ab + 1})] =$

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$\frac{1}{a}$

$\frac{1}{b}$

answer is B.

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Detailed Solution

$Tan [{Tan}^{- 1} (\frac{1}{a + b}) + {Tan}^{- 1} (\frac{b}{a^{2} + ab + 1})]$
$\begin{array}{l} = Tan [{Tan}^{- 1} (\frac{1}{a + b}) + {Tan}^{- 1} (\frac{(a + b) - a}{1 + (a + b) a})] \\ = Tan [c o t^{- 1} (a + b) + {Tan}^{- 1} (a + b) - {Tan}^{1} a] = Tan (\frac{π}{2} - T a n^{- 1} a) = Tan (c o t^{- 1} a) = Tan (T a n^{- 1} \frac{1}{a}) = \frac{1}{a} \end{array}$