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

 Given  { ((A+B=arctan (1/2))),((A−B=arctan (1/3))) :}   then tan A =?

$$\:\mathrm{Given}\:\begin{cases}{\mathrm{A}+\mathrm{B}=\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{A}−\mathrm{B}=\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{3}}}\end{cases} \\ $$$$\:\mathrm{then}\:\mathrm{tan}\:\mathrm{A}\:=? \\ $$

Answered by liberty last updated on 09/Jan/21

 (1)+(2) ⇒ 2A = arctan (1/2)+arctan (1/3)   tan 2A = ((1/2+1/3)/(1−(1/2)×(1/3))) = 1  ⇒((2tan A)/(1−tan^2 A)) = 1 ; tan^2 A+2tan A=1   (tan A+1)^2 =2 ; tan A=−1+(√2)

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

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