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Question Number 128641 by john_santu last updated on 09/Jan/21

$$\mathrm{Given}\{\begin{array}{l}\mathrm{A}+\mathrm{B}=\arctan \frac{1}{2}\\ \mathrm{A}-\mathrm{B}=\arctan \frac{1}{3}\end{array}$$$$\mathrm{then}\tan \mathrm{A}=?$$

Answered by liberty last updated on 09/Jan/21

$$\left(1\right)+\left(2\right)\Rightarrow 2\mathrm{A}=\arctan \frac{1}{2}+\arctan \frac{1}{3}$$$$\tan 2\mathrm{A}=\frac{1/2+1/3}{1-\left(1/2\right)\times \left(1/3\right)}=1$$$$\Rightarrow \frac{2\tan \mathrm{A}}{1-\tan {}^2\mathrm{A}}=1;\tan {}^2\mathrm{A}+2\tan \mathrm{A}=1$$$${(\tan \mathrm{A}+1)}^2=2;\tan \mathrm{A}=-1+\sqrt{2}$$

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