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Question Number 92691 by john santu last updated on 08/May/20

$$\mathrm{find}xineq{\tan }^{-1}\left(x\right)={\cos }^{-1}\left(x\right)$$

Commented by Prithwish Sen 1 last updated on 08/May/20

$$\mathrm{let}{\tan }^{-1}\mathrm{x}=\theta \mathrm{then}\tan \theta =\mathrm{x}$$$$\cos \theta =\frac{1}{\sqrt{1+{\mathrm{x}}^2}}\Rightarrow \mathrm{x}=\frac{1}{\sqrt{1+{\mathrm{x}}^2}}$$$${\mathrm{x}}^4+{\mathrm{x}}^2-1=0$$$${\mathrm{x}}^2=\frac{-1\pm \sqrt{5}}{2}$$$$\therefore \mathrm{x}=\pm \sqrt{\frac{-1\pm \sqrt{5}}{2}}$$$$\because -1-\sqrt{5}<0\therefore \mathrm{x}=\pm \sqrt{\frac{-1+\sqrt{5}}{2}}$$

Commented by mr W last updated on 08/May/20

$$onlyonesolutionx=\sqrt{\frac{-1+\sqrt{5}}{2}}$$$$becauseforx<0:$$$${\cos }^{-1}x>\frac{\pi }{2}and{\tan }^{-1}x<0,{it}^{\prime }simpossible$$$$that{\tan }^{-1}x={\cos }^{-1}x$$

Commented by Ar Brandon last updated on 08/May/20

OK, thanks for the clarification.

Answered by Ar Brandon last updated on 08/May/20

$${\tan }^{-1}\left(\mathrm{x}\right)={\cos }^{-1}\left(\mathrm{x}\right)$$$$\Rightarrow \mathrm{x}=\tan \left({\cos }^{-1}\left(\mathrm{x}\right)\right)=\frac{\sin \left({\cos }^{-1}\left(\mathrm{x}\right)\right)}{\mathrm{x}}$$$$\Rightarrow {\mathrm{x}}^2=\sin \left({\cos }^{-1}\left(\mathrm{x}\right)\right)$$$$\Rightarrow {\mathrm{x}}^2=\sqrt{1-{\left(\cos \left({\cos }^{-1}\left(\mathrm{x}\right)\right)\right)}^2}$$$$\Rightarrow {\mathrm{x}}^2=\sqrt{1-{\mathrm{x}}^2}$$$$\Rightarrow {\mathrm{x}}^4+{\mathrm{x}}^2-1=0$$$$\Rightarrow {\mathrm{x}}^2=\frac{-1\pm \sqrt{5}}{2}-1-\sqrt{5}<0$$$$\Rightarrow {\mathrm{x}}^2=\frac{-1+\sqrt{5}}{2}\Rightarrow \mathrm{x}=\pm \sqrt{\frac{-1+\sqrt{5}}{2}}$$

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