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Question Number 13403 by Tinkutara last updated on 19/May/17

If tan (A − B) = 1, sec (A + B) = (2/(√3)) , then prove that the smallest positive value of B is ((19π)/(24)) .

$$\mathrm{If}\:\mathrm{tan}\:\left({A}\:−\:{B}\right)\:=\:\mathrm{1},\:\mathrm{sec}\:\left({A}\:+\:{B}\right)\:=\:\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:, \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{positive} \\ $$$$\mathrm{value}\:\mathrm{of}\:{B}\:\mathrm{is}\:\frac{\mathrm{19}\pi}{\mathrm{24}}\:. \\ $$

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 (((19π)/(24))) is not true.

$${excuse}\:{me}\:,{but}\:{i}\:{think}\:{your}\:{answer}\:\left(\frac{\mathrm{19}\pi}{\mathrm{24}}\right)\:{is}\:{not}\:{true}. \\ $$$$ \\ $$

Commented by prakash jain last updated on 20/May/17

B=((7π)/(24)) A=((37π)/(24)) A+B=((44π)/(24)) , sec (((44π)/(24)))=(2/(√3)) A−B=((30π)/(24)), tan (((30π)/(24)))=1

$$\mathrm{B}=\frac{\mathrm{7}\pi}{\mathrm{24}} \\ $$$$\mathrm{A}=\frac{\mathrm{37}\pi}{\mathrm{24}} \\ $$$${A}+{B}=\frac{\mathrm{44}\pi}{\mathrm{24}}\:,\:\mathrm{sec}\:\left(\frac{\mathrm{44}\pi}{\mathrm{24}}\right)=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}} \\ $$$${A}−{B}=\frac{\mathrm{30}\pi}{\mathrm{24}},\:\mathrm{tan}\:\left(\frac{\mathrm{30}\pi}{\mathrm{24}}\right)=\mathrm{1} \\ $$

Answered by ajfour last updated on 19/May/17

tan (A−B)=1 ⇒ A−B=(π/4)+nπ ....(i) sec (A+B)=(2/(√3)) ⇒cos (A+B)=((√3)/2) ⇒ A+B=2mπ±(π/6) .....(ii) (ii)−(i) gives: 2B=(2m−n)π−(π/(12)) , or ...(a) 2B=(2m−n)π−((5π)/(12)) ...(b) if B is to be smallest and positive let us take 2m−n=1 in (b) B=((7π)/(24)) .

$$\mathrm{tan}\:\left({A}−{B}\right)=\mathrm{1} \\ $$$$\Rightarrow\:{A}−{B}=\frac{\pi}{\mathrm{4}}+{n}\pi\:\:\:\:\:\:\:\:\:....\left({i}\right) \\ $$$$\mathrm{sec}\:\left({A}+{B}\right)=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:\Rightarrow\mathrm{cos}\:\left({A}+{B}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{A}+{B}=\mathrm{2}{m}\pi\pm\frac{\pi}{\mathrm{6}}\:\:\:\:\:\:.....\left({ii}\right) \\ $$$$\left({ii}\right)−\left({i}\right)\:{gives}: \\ $$$$\mathrm{2}{B}=\left(\mathrm{2}{m}−{n}\right)\pi−\frac{\pi}{\mathrm{12}}\:\:\:\:\:,\:\:{or}\:\:\:...\left({a}\right) \\ $$$$\mathrm{2}{B}=\left(\mathrm{2}{m}−{n}\right)\pi−\frac{\mathrm{5}\pi}{\mathrm{12}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\left({b}\right) \\ $$$${if}\:{B}\:{is}\:{to}\:{be}\:{smallest}\:{and}\:{positive} \\ $$$${let}\:{us}\:{take}\:\:\mathrm{2}{m}−{n}=\mathrm{1}\:\:{in}\:\:\left({b}\right) \\ $$$${B}=\frac{\mathrm{7}\pi}{\mathrm{24}}\:. \\ $$

Commented by ajfour last updated on 19/May/17

someone resolve the conflict please...

$${someone}\:{resolve}\:{the}\:{conflict}\:{please}... \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/May/17

if you put:2m−n=2⇒2B=2π−((5π)/(12))=((19π)/(12)) ⇒B=((19π)/(24))

$${if}\:{you}\:{put}:\mathrm{2}{m}−{n}=\mathrm{2}\Rightarrow\mathrm{2}{B}=\mathrm{2}\pi−\frac{\mathrm{5}\pi}{\mathrm{12}}=\frac{\mathrm{19}\pi}{\mathrm{12}} \\ $$$$\Rightarrow{B}=\frac{\mathrm{19}\pi}{\mathrm{24}} \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/May/17

tg^2 (A+B)=sec^2 (A+B)−1=(4/3)−1=(1/3) ⇒tg(A+B)=±((√3)/3) tg(2B)=((tg(A+B)−tg(A−B))/(1+tg(A+B).tg(A−B)))=((((√3)/3)−1)/(1+((√3)/3)))= =(((√3)−3)/((√3)+3))=((−((√3)−1))/((√3)+1))=((−(4−2(√(3))))/2)=−(2−(√3))=−tg(π/(12)) =tg(π−(π/(12)))=tg((11π)/(12))⇒2B=((11π)/(12))⇒B=((11π)/(24)) .■ tg(2B)=((((−(√3))/3)−1)/(1−((√3)/3)))=(((√3)+3)/((√3)−3))=(((√3)+1)/(1−(√3)))=−(2+(√3)) =−tg(((5π)/(12)))=tg(π−((5π)/(12)))=tg((7π)/(12))⇒B=((7π)/(24)) .■

$${tg}^{\mathrm{2}} \left({A}+{B}\right)={sec}^{\mathrm{2}} \left({A}+{B}\right)−\mathrm{1}=\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{tg}\left({A}+{B}\right)=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${tg}\left(\mathrm{2}{B}\right)=\frac{{tg}\left({A}+{B}\right)−{tg}\left({A}−{B}\right)}{\mathrm{1}+{tg}\left({A}+{B}\right).{tg}\left({A}−{B}\right)}=\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}−\mathrm{1}}{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}= \\ $$$$=\frac{\sqrt{\mathrm{3}}−\mathrm{3}}{\sqrt{\mathrm{3}}+\mathrm{3}}=\frac{−\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\sqrt{\mathrm{3}}+\mathrm{1}}=\frac{−\left(\mathrm{4}−\mathrm{2}\sqrt{\left.\mathrm{3}\right)}\right.}{\mathrm{2}}=−\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)=−{tg}\frac{\pi}{\mathrm{12}} \\ $$$$={tg}\left(\pi−\frac{\pi}{\mathrm{12}}\right)={tg}\frac{\mathrm{11}\pi}{\mathrm{12}}\Rightarrow\mathrm{2}{B}=\frac{\mathrm{11}\pi}{\mathrm{12}}\Rightarrow{B}=\frac{\mathrm{11}\pi}{\mathrm{24}}\:\:.\blacksquare \\ $$$${tg}\left(\mathrm{2}{B}\right)=\frac{\frac{−\sqrt{\mathrm{3}}}{\mathrm{3}}−\mathrm{1}}{\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}=\frac{\sqrt{\mathrm{3}}+\mathrm{3}}{\sqrt{\mathrm{3}}−\mathrm{3}}=\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{1}−\sqrt{\mathrm{3}}}=−\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$$$=−{tg}\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right)={tg}\left(\pi−\frac{\mathrm{5}\pi}{\mathrm{12}}\right)={tg}\frac{\mathrm{7}\pi}{\mathrm{12}}\Rightarrow{B}=\frac{\mathrm{7}\pi}{\mathrm{24}}\:.\blacksquare \\ $$

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