If A>0,B>0, and A+B=(π/3) , then maximum value of tanAtanB is ?


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Question Number 57269 by rahul 19 last updated on 01/Apr/19

$I f A > 0, B > 0, a n d A + B = \frac{π}{3}, t h e n$ $m a x i m u m v a l u e o f t a n A t a n B i s ?$
Commented byrahul 19 last updated on 01/Apr/19

$A n y s h o r t m e t h o d ?$
Commented bytanmay.chaudhury50@gmail.com last updated on 01/Apr/19

Commented bytanmay.chaudhury50@gmail.com last updated on 01/Apr/19

Commented bytanmay.chaudhury50@gmail.com last updated on 01/Apr/19

$t a n A t a n B m a x w h e n A = B = \frac{π}{6}$
Commented bymr W last updated on 01/Apr/19

$d u e t o s y m m e t r y \tan A \times \tan B i s$ $m a x i m u m o r m i n i m u m w h e n A = B = \frac{π}{6},$ $m a x . = \frac{\sqrt{3}}{3} \times \frac{\sqrt{3}}{3} = \frac{1}{3}$
Commented byrahul 19 last updated on 01/Apr/19
thank you sir!
	Answered by tanmay.chaudhury50@gmail.com last updated on 01/Apr/19

	$t a n (A + B) = \sqrt{3}$ $p = t a n A t a n B$ $p = t a n A t a n (\frac{π}{3} - A)$ $p = t a n A \times \frac{\sqrt{3} - t a n A}{1 + \sqrt{3} t a n A}$ $p = \frac{\sqrt{3} a - a^{2}}{1 + \sqrt{3} a}$ $\frac{d p}{d a} = \frac{(1 + \sqrt{3} a) (\sqrt{3} - 2 a) - (\sqrt{3} a - a^{2}) (0 + \sqrt{3})}{{(1 + \sqrt{3} a)}^{2}}$ $f o r m a x / m i n \frac{d p}{d a} = 0$ $\sqrt{3} - 2 a + 3 a - 2 \sqrt{3} a^{2} - 3 a + \sqrt{3} a^{2} = 0$ $- \sqrt{3} a^{2} - 2 a + \sqrt{3} = 0$ $\sqrt{3} a^{2} + 2 a - \sqrt{3} = 0$ $\sqrt{3} a^{2} + 3 a - a - \sqrt{3} = 0$ $\sqrt{3} a (a + \sqrt{3}) - 1 (a + \sqrt{3}) = 0$ $(a + \sqrt{3}) (\sqrt{3} a - 1) = 0$ $a \neq - \sqrt{3} s o a = \frac{1}{\sqrt{3}}$ $\frac{d p}{d a} = \frac{- \sqrt{3} a^{2} - 2 a + \sqrt{3}}{{(1 + \sqrt{3} a)}^{2}}$ $\frac{d p}{d a} > 0 w h e n a < \frac{1}{\sqrt{3}}$ $\frac{d p}{d a} = 0 a t a = \frac{1}{\sqrt{3}}$ $\frac{d p}{d a} < 0 w h e n a > \frac{1}{\sqrt{3}}$ $s o s i g n o f \frac{d p}{d a} c h a n g e s f r o m (+ v e) t o (- v e)$ $s o a t a = \frac{1}{\sqrt{3}} p i s m a x i m u m$ $t a n A = \frac{1}{\sqrt{3}} = t a n \frac{π}{6}$ $A = \frac{π}{6} h e n c e B = \frac{π}{6}$ $m a x v a l u e (t a n \frac{π}{6}) (t a n \frac{π}{6}) = \frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}} = \frac{1}{3}$
	Commented byrahul 19 last updated on 01/Apr/19
	thank you sir!
	Answered by ajfour last updated on 01/Apr/19

	$\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan Atan B} = \sqrt{3}$ $\Rightarrow \tan Atan B = 1 - \frac{\tan A + \tan B}{\sqrt{3}}$ $let f (x) = \tan x + \tan (π / 3 - x)$ $f^{'} (x) = \sec^{2} x - \sec^{2} (π / 3 - x)$ $= 0 \Rightarrow x = \frac{π}{3} - x$ $\Rightarrow x = \frac{π}{6}$ $f^{″} (\frac{π}{6}) = 2 \sec^{2} \frac{π}{6} \tan \frac{π}{6}$ $+ 2 \sec^{2} \frac{π}{6} \tan \frac{π}{6} > 0$ $hence {(\tan Atan B)}_{\max} = \frac{1}{3} .$
	Commented byrahul 19 last updated on 01/Apr/19
	thank you sir!
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