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Question Number 167502 by cortano1 last updated on 18/Mar/22

$$\{\begin{array}{l}\tan \alpha =\frac{\mathrm{m}}{\mathrm{m}+1}\\ \tan \beta =\frac{1}{2\mathrm{m}+1}\end{array}\Rightarrow \alpha +\beta =?$$

Answered by Rasheed.Sindhi last updated on 18/Mar/22

$$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$$$$=\frac{\frac{\mathrm{m}}{\mathrm{m}+1}+\frac{1}{2\mathrm{m}+1}}{1-\frac{\mathrm{m}}{\mathrm{m}+1}\cdot \frac{1}{2\mathrm{m}+1}}=\frac{\mathrm{m}(2\mathrm{m}+1)+\mathrm{m}+1}{(\mathrm{m}+1)(2\mathrm{m}+1)-\mathrm{m}}$$$$=\frac{{2\mathrm{m}}^2+2\mathrm{m}+1}{{2\mathrm{m}}^2+2\mathrm{m}+1}$$$$\alpha +\beta ={\tan }^{-1}\left(1\right)=\frac{\pi }{4}+n\pi$$$$$$

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