Question Number 13403 by Tinkutara last updated on 19/May/17
$$\mathrm{If}\:\mathrm{tan}\:\left({A}\:−\:{B}\right)\:=\:\mathrm{1},\:\mathrm{sec}\:\left({A}\:+\:{B}\right)\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:, \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{positive} \\ $$$$\mathrm{value}\:\mathrm{of}\:{B}\:\mathrm{is}\:\frac{\mathrm{19}\pi}{\mathrm{24}}\:. \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/May/17
$${excuse}\:{me}\:,{but}\:{i}\:{think}\:{your}\:{answer}\:\left(\frac{\mathrm{19}\pi}{\mathrm{24}}\right)\:{is}\:{not}\:{true}. \\ $$$$ \\ $$
Commented by prakash jain last updated on 20/May/17
$$\mathrm{B}=\frac{\mathrm{7}\pi}{\mathrm{24}} \\ $$$$\mathrm{A}=\frac{\mathrm{37}\pi}{\mathrm{24}} \\ $$$${A}+{B}=\frac{\mathrm{44}\pi}{\mathrm{24}}\:,\:\mathrm{sec}\:\left(\frac{\mathrm{44}\pi}{\mathrm{24}}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$$${A}−{B}=\frac{\mathrm{30}\pi}{\mathrm{24}},\:\mathrm{tan}\:\left(\frac{\mathrm{30}\pi}{\mathrm{24}}\right)=\mathrm{1} \\ $$
Answered by ajfour last updated on 19/May/17
$$\mathrm{tan}\:\left({A}−{B}\right)=\mathrm{1} \\ $$$$\Rightarrow\:{A}−{B}=\frac{\pi}{\mathrm{4}}+{n}\pi\:\:\:\:\:\:\:\:\:….\left({i}\right) \\ $$$$\mathrm{sec}\:\left({A}+{B}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\Rightarrow\mathrm{cos}\:\left({A}+{B}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{A}+{B}=\mathrm{2}{m}\pi\pm\frac{\pi}{\mathrm{6}}\:\:\:\:\:\:…..\left({ii}\right) \\ $$$$\left({ii}\right)−\left({i}\right)\:{gives}: \\ $$$$\mathrm{2}{B}=\left(\mathrm{2}{m}−{n}\right)\pi−\frac{\pi}{\mathrm{12}}\:\:\:\:\:,\:\:{or}\:\:\:…\left({a}\right) \\ $$$$\mathrm{2}{B}=\left(\mathrm{2}{m}−{n}\right)\pi−\frac{\mathrm{5}\pi}{\mathrm{12}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({b}\right) \\ $$$${if}\:{B}\:{is}\:{to}\:{be}\:{smallest}\:{and}\:{positive} \\ $$$${let}\:{us}\:{take}\:\:\mathrm{2}{m}−{n}=\mathrm{1}\:\:{in}\:\:\left({b}\right) \\ $$$${B}=\frac{\mathrm{7}\pi}{\mathrm{24}}\:. \\ $$
Commented by ajfour last updated on 19/May/17
$${someone}\:{resolve}\:{the}\:{conflict}\:{please}… \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/May/17
$${if}\:{you}\:{put}:\mathrm{2}{m}−{n}=\mathrm{2}\Rightarrow\mathrm{2}{B}=\mathrm{2}\pi−\frac{\mathrm{5}\pi}{\mathrm{12}}=\frac{\mathrm{19}\pi}{\mathrm{12}} \\ $$$$\Rightarrow{B}=\frac{\mathrm{19}\pi}{\mathrm{24}} \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/May/17
$${tg}^{\mathrm{2}} \left({A}+{B}\right)={sec}^{\mathrm{2}} \left({A}+{B}\right)−\mathrm{1}=\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{tg}\left({A}+{B}\right)=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${tg}\left(\mathrm{2}{B}\right)=\frac{{tg}\left({A}+{B}\right)−{tg}\left({A}−{B}\right)}{\mathrm{1}+{tg}\left({A}+{B}\right).{tg}\left({A}−{B}\right)}=\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}−\mathrm{1}}{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}= \\ $$$$=\frac{\sqrt{\mathrm{3}}−\mathrm{3}}{\:\sqrt{\mathrm{3}}+\mathrm{3}}=\frac{−\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\:\sqrt{\mathrm{3}}+\mathrm{1}}=\frac{−\left(\mathrm{4}−\mathrm{2}\sqrt{\left.\mathrm{3}\right)}\right.}{\mathrm{2}}=−\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)=−{tg}\frac{\pi}{\mathrm{12}} \\ $$$$={tg}\left(\pi−\frac{\pi}{\mathrm{12}}\right)={tg}\frac{\mathrm{11}\pi}{\mathrm{12}}\Rightarrow\mathrm{2}{B}=\frac{\mathrm{11}\pi}{\mathrm{12}}\Rightarrow{B}=\frac{\mathrm{11}\pi}{\mathrm{24}}\:\:.\blacksquare \\ $$$${tg}\left(\mathrm{2}{B}\right)=\frac{\frac{−\sqrt{\mathrm{3}}}{\mathrm{3}}−\mathrm{1}}{\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}=\frac{\sqrt{\mathrm{3}}+\mathrm{3}}{\:\sqrt{\mathrm{3}}−\mathrm{3}}=\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{1}−\sqrt{\mathrm{3}}}=−\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$$$=−{tg}\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right)={tg}\left(\pi−\frac{\mathrm{5}\pi}{\mathrm{12}}\right)={tg}\frac{\mathrm{7}\pi}{\mathrm{12}}\Rightarrow{B}=\frac{\mathrm{7}\pi}{\mathrm{24}}\:.\blacksquare \\ $$