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Question Number 1157 by navajyoti.tamuli.tamuli@gmail. last updated on 06/Jul/15
show that tan^(−1) ((((√(1+x^2 ))−1)/x))=(1/2)tan^(−1) x
$${show}\:{that} \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} {x} \\ $$$$ \\ $$
Answered by prakash jain last updated on 11/Jul/15
Let y=tan^(−1) x⇒x=tany 1+x^2 =sec^2 y (((√(1+x^2 ))−1)/x)=(((√(sec^2 y))−1)/(tany))=((secy−1)/(tany))=((1−cosy)/(siny)) =((1−(1−2sin^2 (y/2)))/(2sin(y/2)cos(y/2)))=tan(y/2) L.H.S=tan^(−1) tan(y/2)=(y/2) R.H.S=(1/2)tan^(−1) x=(y/2) ∴L.H.S=R.H.S
$$\boldsymbol{\mathrm{Let}}\:\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \boldsymbol{\mathrm{x}}\Rightarrow\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{tany}} \\ $$$$\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} =\boldsymbol{\mathrm{sec}}^{\mathrm{2}} \boldsymbol{\mathrm{y}} \\ $$$$\frac{\sqrt{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }−\mathrm{1}}{\boldsymbol{\mathrm{x}}}=\frac{\sqrt{\boldsymbol{\mathrm{sec}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}−\mathrm{1}}{\boldsymbol{\mathrm{tany}}}=\frac{\boldsymbol{\mathrm{secy}}−\mathrm{1}}{\boldsymbol{\mathrm{tany}}}=\frac{\mathrm{1}−\boldsymbol{\mathrm{cosy}}}{\boldsymbol{\mathrm{siny}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}−\left(\mathrm{1}−\mathrm{2}\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \frac{\boldsymbol{\mathrm{y}}}{\mathrm{2}}\right)}{\mathrm{2}\boldsymbol{\mathrm{sin}}\frac{\boldsymbol{\mathrm{y}}}{\mathrm{2}}\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\mathrm{y}}}{\mathrm{2}}}=\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\mathrm{y}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{L}}.\boldsymbol{\mathrm{H}}.\boldsymbol{\mathrm{S}}=\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\mathrm{y}}}{\mathrm{2}}=\frac{\boldsymbol{\mathrm{y}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{R}}.\boldsymbol{\mathrm{H}}.\boldsymbol{\mathrm{S}}=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{y}}}{\mathrm{2}} \\ $$$$\therefore\boldsymbol{\mathrm{L}}.\boldsymbol{\mathrm{H}}.\boldsymbol{\mathrm{S}}=\boldsymbol{\mathrm{R}}.\boldsymbol{\mathrm{H}}.\boldsymbol{\mathrm{S}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

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