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Q.Solve the equation tan⁡−1(x+1)+tan⁡−1(x−1)=tan⁡−1831\tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\dfrac{8}{31}.

Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2018Subjective· 2mImportance★★★★★
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Combining the two inverse tangents gives 4x² + 31x − 8 = 0, whose valid root is x = 1/4.

We solve tan⁻¹(x+1) + tan⁻¹(x−1) = tan⁻¹(8/31).

Step 1: Apply tan⁻¹A + tan⁻¹B = tan⁻¹[(A+B)/(1−AB)] with A = x+1, B = x−1:

A + B = 2x, and AB = (x+1)(x−1) = x² − 1.

So LHS = tan⁻¹[ 2x / (1 − (x²−1)) ] = tan⁻¹[ 2x / (2 − x²) ].

Step 2: Equate arguments: 2x/(2 − x²) = 8/31.

Cross-multiply: 31(2x) = 8(2 − x²) ⇒ 62x = 16 − 8x²

⇒ 8x² + 62x − 16 = 0 ⇒ 4x² + 31x − 8 = 0.

Step 3: Solve: x = [−31 ± √(31² + 4·4·8)] / (2·4) = [−31 ± √1089]/8 = [−31 ± 33]/8. …

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