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Miscellaneous Exercise 3 · Q136

Q.If x,y,zx, y, z are positive then prove that tan⁡−1x−y1+xy+tan⁡−1y−z1+yz+tan⁡−1z−x1+zx=0\tan^{-1}\dfrac{x-y}{1+xy} + \tan^{-1}\dfrac{y-z}{1+yz} + \tan^{-1}\dfrac{z-x}{1+zx} = 0.

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By the tan-difference formula (property xvi), tan⁡−1x−y1+xy=tan⁡−1x−tan⁡−1y\tan^{-1}\dfrac{x-y}{1+xy}=\tan^{-1}x-\tan^{-1}y (for x,y>0x,y>0), and similarly tan⁡−1y−z1+yz=tan⁡−1y−tan⁡−1z\tan^{-1}\dfrac{y-z}{1+yz}=\tan^{-1}y-\tan^{-1}z and tan⁡−1z−x1+zx=tan⁡−1z−tan⁡−1x\tan^{-1}\dfrac{z-x}{1+zx}=\tan^{-1}z-\tan^{-1}x. Adding all three: $(\tan^{-1}x-\tan^{-1}y)+(\tan^{-1}y-\tan^{-1}z)+(\tan^{-1}z-\tan^{- …

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