MCQ
If $A> 0, B > 0$ and $A + B = \frac{\pi }{6}$, then the minimum value of $tan\,A + tan\,B$ is
  • A
    $\sqrt 3  - \sqrt 2 $
  • $ 4  - 2\sqrt 3 $
  • C
    $\frac{2}{{\sqrt 3 }}$
  • D
    $2 - \sqrt 3 $

Answer

Correct option: B.
$ 4  - 2\sqrt 3 $
b
$\tan \,(A + B)\, = \,\frac{{\tan \,A\, + \tan \,B}}{{1 - \tan \,A\,\tan \,B}}$

$ \Rightarrow \,\frac{1}{{\sqrt 3 }}\, = \,\frac{y}{{1 - \tan \,A\,\tan \,B}}$ where $y\, = \tan \,A\, + \tan \,B$

$ \Rightarrow \tan \,A\,\tan \,B\, = \,1 - \sqrt 3 y$ Also $AM \geqslant GM$

$ \Rightarrow \,\frac{{\tan \,A\, + \,\tan \,B}}{2}\, \geqslant \,\sqrt {\tan \,A\,\tan \,B} $

$ \Rightarrow \,y\, \geqslant \,2\sqrt {1 - \sqrt 3 y} $

$ \Rightarrow \,{y^2}\, \geqslant \,4 - 4\sqrt 3 y$

$ \Rightarrow \,{y^2}\, + \,4\sqrt 3 y - 4 \geqslant 0$

$ \Rightarrow \,y\, \leqslant \, - \,2\sqrt 3  - 4$ or $ \Rightarrow \,y\,  \geqslant  \, - \,2\sqrt 3  + 4$

( $y\, \leqslant \, - \,2\sqrt 3  - 4$ is not possible as $\tan A\,\tan B\, > 0$ 

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