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Exercise 4.5 · Q6

Q.If tan⁡−1x+tan⁡−1y+tan⁡−1z=π\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \pi, show that x+y+z=xyzx+y+z = xyz.

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From tan⁡−1x+tan⁡−1y+tan⁡−1z=π\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\pi we isolate two of the three terms, apply tan⁡\tan to both sides, and simplify using the addition formula and tan⁡(π−θ)=−tan⁡θ\tan(\pi-\theta)=-\tan\theta.

Step 1. Set A=tan⁡−1x, B=tan⁡−1y, C=tan⁡−1zA=\tan^{-1}x,\ B=\tan^{-1}y,\ C=\tan^{-1}z, so A+B+C=πA+B+C=\pi.

Step 2. Isolate A+BA+B. A+B=π−CA+B=\pi-C.

Step 3. Apply tan⁡\tan to both sides. tan⁡(A+B)=tan⁡(π−C)=−tan⁡C=−z\tan(A+B)=\tan(\pi-C)=-\tan C=-z. …

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