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Question Number 155898 by Tawa11 last updated on 05/Oct/21

$$\mathrm{Show}\mathrm{that}{\tan }^{-1}\left(\frac{1}{3}\right)+{\sin }^{-1}\left(\frac{1}{3}\right)=\frac{\pi }{4}$$

Answered by immortel last updated on 05/Oct/21

$$L=tan({tan}^{-1}\left(\frac{1}{3}\right)+{sin}^{-1}\left(\frac{1}{3}\right))=\frac{tan\left(arctan\left(\frac{1}{3}\right)\right)+tan\left(arcsin\frac{1}{3}\right)}{1-tan\left(arctan\left(\frac{1}{3}\right)\right)tan\left(arcsin\frac{1}{3}\right)}$$$$ortan\left(arcsinx\right)=\frac{x}{\sqrt{1-x^2}}$$$$donctan\left(arcsin\frac{1}{3}\right)=\frac{\frac{1}{3}}{\sqrt{1-\frac{1}{9}}}$$$$=\frac{1}{\sqrt{8}}=\frac{\sqrt{2}}{4}$$$$ainsiL=\frac{\frac{1}{3}+\frac{\sqrt{2}}{4}}{1-\frac{\sqrt{2}}{12}}$$$$=\frac{(4+3\sqrt{2})}{12}\times \frac{12}{12-\sqrt{2}}$$$$=\frac{(4+3\sqrt{2})(12+\sqrt{2})}{(12-\sqrt{2})(12+\sqrt{2})}$$$$=\frac{48+4\sqrt{2}+36\sqrt{2}+6}{144-2}$$$$=\frac{54+40\sqrt{2}}{142}=\frac{27+20\sqrt{2}}{71}<1$$$$doncL\ne 1ainsi$$$${\tan }^{-1}\left(\frac{1}{3}\right)+{\sin }^{-1}\left(\frac{1}{3}\right)\ne \frac{\pi }{4}$$$$$$$$..........l^{\prime }immortel........$$$$$$$$$$

Commented by Tawa11 last updated on 05/Oct/21

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by mr W last updated on 05/Oct/21

$$\alpha ={\tan }^{-1}\frac{1}{3}\Rightarrow \tan \alpha =\frac{1}{3}$$$$\beta ={\sin }^{-1}\frac{1}{3}\Rightarrow \sin \beta =\frac{1}{3}\Rightarrow \tan \beta =\frac{1}{2\sqrt{2}}$$$$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$$$$=\frac{\frac{1}{3}+\frac{1}{2\sqrt{2}}}{1-\frac{1}{6\sqrt{2}}}=\frac{3+2\sqrt{2}}{6\sqrt{2}-1}=\frac{27+20\sqrt{2}}{71}$$$$\approx 0.778\ne 1$$$$\Rightarrow \alpha +\beta \ne 45°$$$$infact{\tan }^{-1}\left(\frac{1}{3}\right)+{\sin }^{-1}\left(\frac{1}{3}\right)\approx 38°$$

Commented by Tawa11 last updated on 05/Oct/21

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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