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Question Number 113200 by bobhans last updated on 11/Sep/20

prove that 2tan^(−1) ((1/3))+tan^(−1) ((1/7))=(π/4)

$$\mathrm{prove}\:\mathrm{that}\:\mathrm{2tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{7}}\right)=\frac{\pi}{\mathrm{4}} \\ $$

Answered by john santu last updated on 11/Sep/20

(Q) prove that 2 tan^(−1) ((1/3))+tan^(−1) ((1/7)) = (π/4). (sol) recall tan^(−1) ((1/3)) = arg (3+i) → 2 tan^(−1) ((1/3)) = 2 arg (3+i) = arg((3+i)^2 )=arg(8+6i) = arg (4+3i) now we have 2 tan^(−1) ((1/3))+tan^(−1) ((1/7)) = arg(4+3i) + arg(7+i) = arg ((4+3i)(7+i)) = arg (25+25i) = arg(1+i) = (π/4).(✓)

$$\left({Q}\right)\:{prove}\:{that}\:\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{7}}\right)\:=\:\frac{\pi}{\mathrm{4}}. \\ $$$$\left({sol}\right)\:{recall}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\:{arg}\:\left(\mathrm{3}+{i}\right) \\ $$$$\rightarrow\:\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\:\mathrm{2}\:{arg}\:\left(\mathrm{3}+{i}\right)\:=\:{arg}\left(\left(\mathrm{3}+{i}\right)^{\mathrm{2}} \right)={arg}\left(\mathrm{8}+\mathrm{6}{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{arg}\:\left(\mathrm{4}+\mathrm{3}{i}\right) \\ $$$${now}\:{we}\:{have}\:\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{7}}\right)\:= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{arg}\left(\mathrm{4}+\mathrm{3}{i}\right)\:+\:{arg}\left(\mathrm{7}+{i}\right)\:= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{arg}\:\left(\left(\mathrm{4}+\mathrm{3}{i}\right)\left(\mathrm{7}+{i}\right)\right)\:=\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{arg}\:\left(\mathrm{25}+\mathrm{25}{i}\right)\:=\:{arg}\left(\mathrm{1}+{i}\right)\:=\:\frac{\pi}{\mathrm{4}}.\left(\checkmark\right) \\ $$

Answered by Dwaipayan Shikari last updated on 11/Sep/20

2tan^(−1) (1/3)=tan^(−1) ((((1/3)+(1/3))/(1−(1/9))))=tan^(−1) (3/4) tan^(−1) ((1/7))+tan^(−1) ((3/4))=tan^(−1) ((((1/7)+(3/4))/(1−(3/(28)))))=tan^(−1) (1)=(π/4)

$$\mathrm{2}{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}={tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{9}}}\right)={tan}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}} \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{7}}\right)+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)={tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{28}}}\right)={tan}^{−\mathrm{1}} \left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}} \\ $$

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