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

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detailed solution

Correct option is B

we have tantan−11a+b+tan−1ba2+ab+1 =tantan−11a+b+tan−1(a+b)−a1+(a+b)a =tantan−11a+b+tan−1(a+b)-tan−1a =tanπ2-tan−1a =tancot−1a =tantan−11a =1a

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tantan−11a+b+tan−1ba2+ab+1=

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