Question

If $A > 0,B > 0$ and $A + B = \dfrac{\pi }{6}$, then the minimum value of tan A + tan B is:
$\left( a \right)\sqrt 3 - \sqrt 2 $
$\left( b \right)4 - 2\sqrt 3 $
$\left( c \right)\dfrac{2}{{\sqrt 3 }}$
$\left( d \right)2 - \sqrt 3 $

Answer 1

Hint: In this particular question use the concept that tan x = (sin x/cos x), $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$ and use the standard angle value of sin x and cos x, i.e.$\sin {30^o} = \dfrac{1}{2}$ and $\cos {30^o} = \dfrac{{\sqrt 3 }}{2}$, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given data:
If $A > 0,B > 0$
And, $A + B = \dfrac{\pi }{6}$............... (1)
Then we have to find out the minimum value of tan A + tan B.
$ \Rightarrow \tan A + \tan B = \dfrac{{\sin A}}{{\cos A}} + \dfrac{{\sin B}}{{\cos B}}$, $\left[ {\because \tan x = \dfrac{{\sin x}}{{\cos x}}} \right]$
$ \Rightarrow \tan A + \tan B = \dfrac{{\sin A\cos B + \cos A\sin B}}{{\cos A\cos B}}$
Now as we know that $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$, so we have,
$ \Rightarrow \tan A + \tan B = \dfrac{{\sin \left( {A + B} \right)}}{{\cos A\cos B}}$
Now divide and multiply by 2 we have,
$ \Rightarrow \tan A + \tan B = \dfrac{{2\sin \left( {A + B} \right)}}{{2\cos A\cos B}}$
Now as we know that 2cos A cos B = cos (A + B) + cos (A – B) so we have,
$ \Rightarrow \tan A + \tan B = \dfrac{{2\sin \left( {A + B} \right)}}{{\cos \left( {A + B} \right) + \cos \left( {A - B} \right)}}$
Now it is given that, $A + B = \dfrac{\pi }{6}$
And we all know that $\dfrac{\pi }{6} = {30^o}$
$ \Rightarrow A + B = {30^o}$
$ \Rightarrow \tan A + \tan B = \dfrac{{2\sin {{30}^o}}}{{\cos {{30}^o} + \cos \left( {A - B} \right)}}$
Now we have to find out the minimum value of tan A + tan B,
So for that we have to maximize the denominator.
So as we know that maximum value of cos x = 1, for x = 0.
So, A – B = 0 for the minimum value of tan A + tan B.
So, cos (A – B) = cos 0 = 0
$ \Rightarrow {\left( {\tan A + \tan B} \right)_{\min }} = \dfrac{{2\sin {{30}^o}}}{{\cos {{30}^o} + \cos \left( {{0^o}} \right)}}$
$ \Rightarrow {\left( {\tan A + \tan B} \right)_{\min }} = \dfrac{{2 \times \dfrac{1}{2}}}{{\dfrac{{\sqrt 3 }}{2} + 1}}$, [as, $\sin {30^o} = \dfrac{1}{2}$ and $\cos {30^o} = \dfrac{{\sqrt 3 }}{2}$]
$ \Rightarrow {\left( {\tan A + \tan B} \right)_{\min }} = \dfrac{2}{{\sqrt 3 + 2}}$
Now rationalize the above equation by divide and multiply by $\left( {\sqrt 3 - 2} \right)$ we have,
\[ \Rightarrow {\left( {\tan A + \tan B} \right)_{\min }} = \dfrac{2}{{\sqrt 3 + 2}} \times \dfrac{{\sqrt 3 - 2}}{{\sqrt 3 - 2}}\]
\[ \Rightarrow {\left( {\tan A + \tan B} \right)_{\min }} = \dfrac{{2\sqrt 3 - 4}}{{{{\left( {\sqrt 3 } \right)}^2} - {2^2}}}\], $\left[ {\because \left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}} \right]$
\[ \Rightarrow {\left( {\tan A + \tan B} \right)_{\min }} = \dfrac{{2\sqrt 3 - 4}}{{3 - 4}} = \dfrac{{2\sqrt 3 - 4}}{{ - 1}} = 4 - 2\sqrt 3 \]
So this is the required minimum value.

So, the correct answer is “Option (b)”.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic trigonometric identity such as 2cos A cos B = cos (A + B) + cos (A – B), and also recall the standard angle values which are all stated above, then first simplify the given function then apply the basic trigonometric identities as above, then for the minimum value of the function we have to maximize the denominator as above an simplify we will get the required answer.