Question Number 24359 by math solver last updated on 16/Nov/17
Commented by ajfour last updated on 17/Nov/17
![If Σ_(r=1) ^(10) tan^(−1) ((3/(9r^2 +3r−1))) =cot^(−1) ((m/n)) , then it means ⇒ Σ_(r=1) ^(10) tan^(−1) [(((3r+2)+(1−3r))/(1−(3r+2)(1−3r)))] =cot^(−1) ((m/n)) or Σ_(r=1) ^(10) [tan^(−1) (3r+2)+tan^(−1) (1−3r)] =cot^(−1) ((m/n)) ⇒ tan^(−1) 5+tan^(−1) (−2) + tan^(−1) 8+tan^(−1) (−5) +...... +tan^(−1) (32)+tan^(−1) (−29) =cot^(−1) ((m/n)) ⇒ tan^(−1) (32)−tan^(−1) 2 =cot^(−1) ((m/n)) tan^(−1) (((32−2)/(1+2×32)))=cot^(−1) ((m/n)) tan^(−1) ((6/(13)))=cot^(−1) ((m/n)) cot^(−1) (((13)/6)) = cot^(−1) ((m/n)) as m, n are coprime to one another , we can infer m=13 , n=6 (13, 6) does lie on the curves in option (A), (C), and (D). It doesn′t lie on curve given in option (B).](Q24399.png)
$${If}\:\underset{{r}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{9}{r}^{\mathrm{2}} +\mathrm{3}{r}−\mathrm{1}}\right) \\ $$$$\:\:\:\:=\mathrm{cot}^{−\mathrm{1}} \left(\frac{{m}}{{n}}\right)\:,\:{then}\:{it}\:{means} \\ $$$$\Rightarrow\:\underset{{r}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left[\frac{\left(\mathrm{3}{r}+\mathrm{2}\right)+\left(\mathrm{1}−\mathrm{3}{r}\right)}{\mathrm{1}−\left(\mathrm{3}{r}+\mathrm{2}\right)\left(\mathrm{1}−\mathrm{3}{r}\right)}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{cot}^{−\mathrm{1}} \left(\frac{{m}}{{n}}\right) \\ $$$${or}\:\underset{{r}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3}{r}+\mathrm{2}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}−\mathrm{3}{r}\right)\right] \\ $$$$\:\:\:\:\:=\mathrm{cot}^{−\mathrm{1}} \left(\frac{{m}}{{n}}\right) \\ $$$$\Rightarrow\:\:\:\:\mathrm{tan}^{−\mathrm{1}} \mathrm{5}+\mathrm{tan}^{−\mathrm{1}} \left(−\mathrm{2}\right) \\ $$$$\:\:\:\:+\:\mathrm{tan}^{−\mathrm{1}} \mathrm{8}+\mathrm{tan}^{−\mathrm{1}} \left(−\mathrm{5}\right) \\ $$$$\:\:\:\:+...... \\ $$$$\:\:\:\:\:+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{32}\right)+\mathrm{tan}^{−\mathrm{1}} \left(−\mathrm{29}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{cot}^{−\mathrm{1}} \left(\frac{{m}}{{n}}\right) \\ $$$$\Rightarrow\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{32}\right)−\mathrm{tan}^{−\mathrm{1}} \mathrm{2}\:=\mathrm{cot}^{−\mathrm{1}} \left(\frac{{m}}{{n}}\right) \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{32}−\mathrm{2}}{\mathrm{1}+\mathrm{2}×\mathrm{32}}\right)=\mathrm{cot}^{−\mathrm{1}} \left(\frac{{m}}{{n}}\right) \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{6}}{\mathrm{13}}\right)=\mathrm{cot}^{−\mathrm{1}} \left(\frac{{m}}{{n}}\right) \\ $$$$\mathrm{cot}^{−\mathrm{1}} \left(\frac{\mathrm{13}}{\mathrm{6}}\right)\:=\:\mathrm{cot}^{−\mathrm{1}} \left(\frac{{m}}{{n}}\right) \\ $$$${as}\:{m},\:{n}\:{are}\:{coprime}\:{to}\:{one}\: \\ $$$${another}\:,\:{we}\:{can}\:{infer} \\ $$$$\:\:\:\:\boldsymbol{{m}}=\mathrm{13}\:,\:\boldsymbol{{n}}=\mathrm{6} \\ $$$$\left(\mathrm{13},\:\mathrm{6}\right)\:{does}\:{lie}\:{on}\:{the}\:{curves} \\ $$$${in}\:{option}\:\left({A}\right),\:\left({C}\right),\:{and}\:\left({D}\right). \\ $$$${It}\:{doesn}'{t}\:{lie}\:{on}\:{curve}\:{given}\:{in} \\ $$$${option}\:\left({B}\right). \\ $$
Answered by ajfour last updated on 17/Nov/17
$$\left({A}\right),\:\left({C}\right),\:{and}\:\left({D}\right) \\ $$
Commented by math solver last updated on 17/Nov/17
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}! \\ $$