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Miscellaneous Exercise 3 · Q123

Q.If tan⁡−1x−1x−2+tan⁡−1x+1x+2=π4\tan^{-1}\dfrac{x-1}{x-2} + \tan^{-1}\dfrac{x+1}{x+2} = \dfrac{\pi}{4} then find the value of xx.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Combining via the tan-sum formula and setting the result equal to tan⁡(π/4)=1\tan(\pi/4)=1: x−1x−2+x+1x+21−(x−1)(x+1)(x−2)(x+2)=1\dfrac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{(x-1)(x+1)}{(x-2)(x+2)}}=1. Numerator: (x−1)(x+2)+(x+1)(x−2)(x−2)(x+2)=(x2+x−2)+(x2−x−2)x2−4=2x2−4x2−4\dfrac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)}=\dfrac{(x^2+x-2)+(x^2-x-2)}{x^2-4}=\dfrac{2x^2-4}{x^2-4}. Denominator: (x2−4)−(x2−1)x2−4=−3x2−4\dfrac{(x^2-4)-(x^2-1)}{x^2-4}=\dfrac{-3}{x^2-4}. Rat …

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