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If-tan-A-B-1-sec-A-B-2-3-then-prove-that-the-smallest-positive-value-of-B-is-19pi-24-




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)) .
Iftan(AB)=1,sec(A+B)=23,thenprovethatthesmallestpositivevalueofBis19π24.
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.
excuseme,butithinkyouranswer(19π24)isnottrue.
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
B=7π24A=37π24A+B=44π24,sec(44π24)=23AB=30π24,tan(30π24)=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)) .
tan(AB)=1AB=π4+nπ.(i)sec(A+B)=23cos(A+B)=32A+B=2mπ±π6..(ii)(ii)(i)gives:2B=(2mn)ππ12,or(a)2B=(2mn)π5π12(b)ifBistobesmallestandpositiveletustake2mn=1in(b)B=7π24.
Commented by ajfour last updated on 19/May/17
someone resolve the conflict please...
someoneresolvetheconflictplease
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))
ifyouput:2mn=22B=2π5π12=19π12B=19π24
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)) .■
tg2(A+B)=sec2(A+B)1=431=13tg(A+B)=±33tg(2B)=tg(A+B)tg(AB)1+tg(A+B).tg(AB)=3311+33==333+3=(31)3+1=(423)2=(23)=tgπ12=tg(ππ12)=tg11π122B=11π12B=11π24.◼tg(2B)=331133=3+333=3+113=(2+3)=tg(5π12)=tg(π5π12)=tg7π12B=7π24.◼

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