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Question Number 94241 by Ar Brandon last updated on 17/May/20

$$\mathrm{Prove}\mathrm{that}$$$$\arctan \left(\mathrm{x}\right)+2\arctan (\sqrt{1+{\mathrm{x}}^2}-\mathrm{x})=\frac{\pi }{2}$$

Commented by mathmax by abdo last updated on 17/May/20

$$let\phi \left(x\right)=arctanx+2srctan(\sqrt{1+x^2}-x)\Rightarrow$$$${\phi }^{{}^{\prime }}\left(x\right)=\frac{1}{1+x^2}+2\times \frac{{(\sqrt{1+x^2}-x)}^{{}^{\prime }}}{1+{(\sqrt{1+x^2}-x)}^2}$$$$=\frac{1}{1+x^2}+2\times \frac{\frac{x}{\sqrt{1+x^2}}-1}{1+1+x^2-2x\sqrt{1+x^2}+x^2}$$$$=\frac{1}{1+x^2}+2\times \frac{x-\sqrt{1+x^2}}{\sqrt{1+x^2}(2+2x^2-2x\sqrt{1+x^2})}$$$$=\frac{1}{1+x^2}+\frac{x-\sqrt{1+x^2}}{\sqrt{1+x^2}(1+x^2-x\sqrt{1+x^2})}$$$$=\frac{(1+x^2)\sqrt{1+x^2}-x(1+x^2)+(1+x^2)x-(1+x^2)\sqrt{1+x^2}}{(1+x^2)\sqrt{1+x^2}(1+x^2-x\sqrt{1+x^2})}=0\Rightarrow$$$$\phi \left(x\right)=c$$$$c=\phi \left(0\right)=0+2arctan\left(1\right)=2\times \frac{\pi }{4}=\frac{\pi }{2}theequalityisproved.$$

Answered by som(math1967) last updated on 17/May/20

$${\tan }^{-1}\mathrm{x}+{2\tan }^{-1}(\sqrt{1+{\mathrm{x}}^2}-\mathrm{x})$$$${\tan }^{-1}\mathrm{x}+{\tan }^{-1}\frac{2(\sqrt{1+{\mathrm{x}}^2}-\mathrm{x})}{1-{(\sqrt{1+{\mathrm{x}}^2}-\mathrm{x})}^2}$$$${\tan }^{-1}\mathrm{x}+{\tan }^{-1}\frac{2(\sqrt{1+{\mathrm{x}}^2}-\mathrm{x})}{1-1-{\mathrm{x}}^2-{\mathrm{x}}^2+2\mathrm{x}\sqrt{1+{\mathrm{x}}^2}}$$$${\tan }^{-1}\mathrm{x}+{\tan }^{-1}\frac{2(\sqrt{1+{\mathrm{x}}^2}-\mathrm{x})}{2\mathrm{x}(\sqrt{1+{\mathrm{x}}^2}-\mathrm{x})}$$$${\tan }^{-1}\mathrm{x}+{\tan }^{-1}\frac{1}{\mathrm{x}}$$$${\tan }^{-1}\mathrm{x}+{\cot }^{-1}\mathrm{x}$$$$=\frac{\pi }{2}$$

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