If-A-gt-0-B-gt-0-and-A-B-pi-3-then-maximum-value-of-tanAtanB-is-

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 by rahul 19 last updated on 01/Apr/19

A n y s h o r t m e t h o d ?

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

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

Commented by tanmay.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 by mr 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 by rahul 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 by rahul 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 by rahul 19 last updated on 01/Apr/19

thank you sir!