If tan-1 2x+tan-1 3x=π4, then x=

tan-12x+3x1-6x=π4⇒x+16x-1=0⇒x=-1or16⇒x=16

If tan-1 2x+tan-1 3x=π4, then x=

tan-12x+3x1-6x=π4⇒x+16x-1=0⇒x=-1or16⇒x=16

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If $\tan^{- 1} 2 x + \tan^{- 1} 3 x = \frac{π}{4},$ then $x =$

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1/6

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-1/6

answer is C.

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Detailed Solution

$\tan^{- 1} (\frac{2 x + 3 x}{1 - 6 x}) = \frac{π}{4} \Rightarrow (x + 1) (6 x - 1) = 0 \Rightarrow x = - 1 (or) \frac{1}{6} \Rightarrow x = \frac{1}{6}$

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If $\tan^{- 1} 2 x + \tan^{- 1} 3 x = \frac{π}{4},$ then $x =$