If tan−1 3 + tan−1 x = tan−1 8, then x =

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(b) $\frac{1}{5}$ We know that \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} x + y/1 - xy Now, \tan^{- 1} 3 + \tan^{- 1} x = \tan^{- 1} 8 \Rightarrow \tan^{- 1} \left( 3 + x/1 - 3x \right) = \tan^{- 1} 8 \Rightarrow 3 + x/1 - 3x = 8 \Rightarrow 3 + x = 8 - 24x \Rightarrow 3 - 8 = - 24x - x \Rightarrow - 5 = - 25x \Rightarrow x = $\frac{5}{25}$ = $\frac{1}{5}$

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