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Question Number 53924 by F_Nongue last updated on 27/Jan/19

$$\mathrm{If}\mathrm{the}\mathrm{non}-\mathrm{zero}\mathrm{numbers}x,y,z\mathrm{are}\mathrm{in}\mathrm{AP},$$$$\mathrm{and}{\tan }^{-1}x,{\tan }^{-1}y,{\tan }^{-1}z\mathrm{are}\mathrm{also}\mathrm{in}\mathrm{AP},$$$$\mathrm{then}$$

Answered by $@ty@m last updated on 27/Jan/19

$${\tan }^{-1}y-{\tan }^{-1}x={\tan }^{-1}z-{\tan }^{-1}y$$$$\Rightarrow {\tan }^{-1}\frac{y-x}{1+yx}={\tan }^{-1}\frac{z-y}{1+zy}$$$$\Rightarrow \frac{y-x}{1+yx}=\frac{z-y}{1+zy}\Rightarrow$$$$\Rightarrow 1+yx=1+zy(\because x,y,z\mathrm{are}\mathrm{in}\mathrm{AP})$$$$\Rightarrow x=z$$$$\Rightarrow y=z$$$$\Rightarrow x=y=z$$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jan/19

$$given2y=x+z$$$$2{tan}^{-1}y={tan}^{-1}x+{tan}^{-1}z$$$${tan}^{-1}\left(\frac{2y}{1-y^2}\right)={tan}^{-1}\left(\frac{x+z}{1-xz}\right)$$$$\frac{2y}{1-y^2}=\frac{x+z}{1-xz}$$$$2y-2xyz=x+z-{xy}^2-{zy}^2$$$${xy}^2+{zy}^2-2xyz=0$$$$xy+zy-2xz=0$$$$\frac{1}{z}+\frac{1}{x}=\frac{2}{y}$$$$x,yandzinH.Palso...$$$$Peculiarconclusion...$$$$plsclearlymentioninquestionwhattoprove...$$$$$$

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