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Miscellaneous Exercise 3 · Q79

Q.If tan⁡−1(2x)+tan⁡−1(3x)=π4\tan^{-1}(2x) + \tan^{-1}(3x) = \dfrac{\pi}{4}, then x=x =

(a) −1-1
(b) 16\dfrac{1}{6}
(c) 26\dfrac{2}{6}
(d) 32\dfrac{3}{2}
Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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tan⁡(π4)=1=2x+3x1−6x2=5x1−6x2  ⟹  1−6x2=5x  ⟹  6x2+5x−1=0  ⟹  (6x−1)(x+1)=0  ⟹  x=16\tan\left(\dfrac{\pi}{4}\right)=1=\dfrac{2x+3x}{1-6x^2}=\dfrac{5x}{1-6x^2}\implies1-6x^2=5x\implies6x^2+5x-1=0\implies(6x-1)(x+1)=0\implies x=\dfrac16 or x=−1x=-1. Checking validity (the sum must stay within the branch where the direct tan-sum formula applies): x=16x=\dfrac16 gives both $2x,3 …

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