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


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

$If \tan (A - B) = 1, \sec (A + B) = \frac{2}{\sqrt{3}},$ $then prove that the smallest positive$ $value of B is \frac{19 π}{24} .$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/May/17

$e x c u s e m e, b u t i t h i n k y o u r a n s w e r (\frac{19 π}{24}) i s n o t t r u e .$
Commented by prakash jain last updated on 20/May/17

$B = \frac{7 π}{24}$ $A = \frac{37 π}{24}$ $A + B = \frac{44 π}{24}, \sec (\frac{44 π}{24}) = \frac{2}{\sqrt{3}}$ $A - B = \frac{30 π}{24}, \tan (\frac{30 π}{24}) = 1$
	Answered by ajfour last updated on 19/May/17

	$\tan (A - B) = 1$ $\Rightarrow A - B = \frac{π}{4} + n π . . . . (i)$ $\sec (A + B) = \frac{2}{\sqrt{3}} \Rightarrow \cos (A + B) = \frac{\sqrt{3}}{2}$ $\Rightarrow A + B = 2 m π \pm \frac{π}{6} . . . . . (i i)$ $(i i) - (i) g i v e s :$ $2 B = (2 m - n) π - \frac{π}{12}, o r . . . (a)$ $2 B = (2 m - n) π - \frac{5 π}{12} . . . (b)$ $i f B i s t o b e s m a l l e s t a n d p o s i t i v e$ $l e t u s t a k e 2 m - n = 1 i n (b)$ $B = \frac{7 π}{24} .$
	Commented by ajfour last updated on 19/May/17

	$s o m e o n e r e s o l v e t h e c o n f l i c t p l e a s e . . .$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/May/17

	$i f y o u p u t : 2 m - n = 2 \Rightarrow 2 B = 2 π - \frac{5 π}{12} = \frac{19 π}{12}$ $\Rightarrow B = \frac{19 π}{24}$
	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/May/17

	${t g}^{2} (A + B) = {s e c}^{2} (A + B) - 1 = \frac{4}{3} - 1 = \frac{1}{3}$ $\Rightarrow t g (A + B) = \pm \frac{\sqrt{3}}{3}$ $t g (2 B) = \frac{t g (A + B) - t g (A - B)}{1 + t g (A + B) . t g (A - B)} = \frac{\frac{\sqrt{3}}{3} - 1}{1 + \frac{\sqrt{3}}{3}} =$ $= \frac{\sqrt{3} - 3}{\sqrt{3} + 3} = \frac{- (\sqrt{3} - 1)}{\sqrt{3} + 1} = \frac{- (4 - 2 \sqrt{3)}}{2} = - (2 - \sqrt{3}) = - t g \frac{π}{12}$ $= t g (π - \frac{π}{12}) = t g \frac{11 π}{12} \Rightarrow 2 B = \frac{11 π}{12} \Rightarrow B = \frac{11 π}{24} . ◼$ $t g (2 B) = \frac{\frac{- \sqrt{3}}{3} - 1}{1 - \frac{\sqrt{3}}{3}} = \frac{\sqrt{3} + 3}{\sqrt{3} - 3} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} = - (2 + \sqrt{3})$ $= - t g (\frac{5 π}{12}) = t g (π - \frac{5 π}{12}) = t g \frac{7 π}{12} \Rightarrow B = \frac{7 π}{24} . ◼$
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