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Question 173 of 177

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

(a) -1
(b) 16\dfrac16
(c) 13\dfrac13
(d) 32\dfrac32
Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2026MCQ· 2mImportance★★★★★
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Use tan⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB)\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left(\dfrac{A+B}{1-AB}\right) and check both roots for validity.

tan⁡−1(2x)+tan⁡−1(3x)=π4\tan^{-1}(2x)+\tan^{-1}(3x)=\frac{\pi}{4}

Taking tangent of both sides:

2x+3x1−(2x)(3x)=tan⁡π4=1\frac{2x+3x}{1-(2x)(3x)}=\tan\frac{\pi}{4}=1

5x1−6x2=1  ⟹  5x=1−6x2  ⟹  6x2+5x−1=0\frac{5x}{1-6x^2}=1 \implies 5x=1-6x^2 \implies 6x^2+5x-1=0

(6x−1)(x+1)=0  ⟹  x=16 or x=−1(6x-1)(x+1)=0 \implies x=\frac16 \text{ or } x=-1

Check validity (need 2x,3x2x,3x such that the sum genuinely equals π/4\pi/4, i.e. 1−6x2>01-6x^2>0): …

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