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Question Number 1157 by navajyoti.tamuli.tamuli@gmail. last updated on 06/Jul/15

$$showthat$$$${tan}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\frac{1}{2}{tan}^{-1}x$$$$$$

Answered by prakash jain last updated on 11/Jul/15

$$\mathbf{Let}\mathbf{y}={\mathbf{tan}}^{-1}\mathbf{x}\Rightarrow \mathbf{x}=\mathbf{tany}$$$$1+{\mathbf{x}}^2={\mathbf{sec}}^2\mathbf{y}$$$$\frac{\sqrt{1+{\mathbf{x}}^2}-1}{\mathbf{x}}=\frac{\sqrt{{\mathbf{sec}}^2\mathbf{y}}-1}{\mathbf{tany}}=\frac{\mathbf{secy}-1}{\mathbf{tany}}=\frac{1-\mathbf{cosy}}{\mathbf{siny}}$$$$=\frac{1-(1-2{\mathbf{sin}}^2\frac{\mathbf{y}}{2})}{2\mathbf{sin}\frac{\mathbf{y}}{2}\mathbf{cos}\frac{\mathbf{y}}{2}}=\mathbf{tan}\frac{\mathbf{y}}{2}$$$$\mathbf{L}.\mathbf{H}.\mathbf{S}={\mathbf{tan}}^{-1}\mathbf{tan}\frac{\mathbf{y}}{2}=\frac{\mathbf{y}}{2}$$$$\mathbf{R}.\mathbf{H}.\mathbf{S}=\frac{1}{2}{\mathbf{tan}}^{-1}\mathbf{x}=\frac{\mathbf{y}}{2}$$$$\therefore \mathbf{L}.\mathbf{H}.\mathbf{S}=\mathbf{R}.\mathbf{H}.\mathbf{S}$$$$$$

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