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Question Number 125186 by Mathgreat last updated on 08/Dec/20

Answered by bemath last updated on 08/Dec/20

$$\left(1\right)\mathrm{arc}\tan \left(\frac{1}{3}\right)+\mathrm{arc}\tan \left(\frac{1}{9}\right)=$$$$\mathrm{arc}\tan \left(\frac{\frac{1}{3}+\frac{1}{9}}{1-\frac{1}{27}}\right)=\mathrm{arc}\tan \left(\frac{12}{26}\right)=\arctan \left(\frac{6}{13}\right)$$$$\left(2\right)\mathrm{arc}\tan \left(\frac{6}{13}\right)+\mathrm{arc}\tan \left(\frac{7}{19}\right)=$$$$\mathrm{arc}\tan \left(\frac{\frac{6}{13}+\frac{7}{19}}{1-\frac{6}{13}.\frac{7}{19}}\right)=\mathrm{arc}\tan \left(\frac{205}{205}\right)=\frac{\pi }{4}$$

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