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Question Number 116966 by bobhans last updated on 08/Oct/20

4 tan^(−1) ((1/5))−tan^(−1) ((1/(239))) = ?

$$\mathrm{4}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{239}}\right)\:=\:? \\ $$

Answered by bemath last updated on 08/Oct/20

Answered by TANMAY PANACEA last updated on 08/Oct/20

2tan^(−1) ((((1/5)+(1/5))/(1−(1/5)×(1/5))))−tan^(−1) ((1/(239)))  =2tan^(−1) (((2×25)/(5×24)))−tan^(−1) ((1/(239)))  =tan^(−1) ((((5/(12))+(5/(12)))/(1−((25)/(144)))))−tan^(−1) ((1/(239)))  =tan^(−1) (((10)/(12))×((144)/(119)))−tan^(−1) ((1/(239)))  =tan^(−1) (((((120)/(119))−(1/(239)))/(1+((120×1)/(119×239)))))=tan^(−1) (((120×239−119)/(119×239+120)))  =tan^(−1) (((119×239+239−119)/(119×239+120)))=tan^(−1) (((119×239+120)/(119×239+120)))  =tan^(−1) (1)=(π/4)

$$\mathrm{2}{tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}×\frac{\mathrm{1}}{\mathrm{5}}}\right)−{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{239}}\right) \\ $$$$=\mathrm{2}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}×\mathrm{25}}{\mathrm{5}×\mathrm{24}}\right)−{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{239}}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{5}}{\mathrm{12}}+\frac{\mathrm{5}}{\mathrm{12}}}{\mathrm{1}−\frac{\mathrm{25}}{\mathrm{144}}}\right)−{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{239}}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\mathrm{10}}{\mathrm{12}}×\frac{\mathrm{144}}{\mathrm{119}}\right)−{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{239}}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{120}}{\mathrm{119}}−\frac{\mathrm{1}}{\mathrm{239}}}{\mathrm{1}+\frac{\mathrm{120}×\mathrm{1}}{\mathrm{119}×\mathrm{239}}}\right)={tan}^{−\mathrm{1}} \left(\frac{\mathrm{120}×\mathrm{239}−\mathrm{119}}{\mathrm{119}×\mathrm{239}+\mathrm{120}}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\mathrm{119}×\mathrm{239}+\mathrm{239}−\mathrm{119}}{\mathrm{119}×\mathrm{239}+\mathrm{120}}\right)={tan}^{−\mathrm{1}} \left(\frac{\mathrm{119}×\mathrm{239}+\mathrm{120}}{\mathrm{119}×\mathrm{239}+\mathrm{120}}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}} \\ $$

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