Question

If A > 0, B > 0 and A + B = $\frac{\pi}{3}$ then the maximum value of tan A . tan B is –

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{1}{3}$
• C. 3
• D. $\sqrt{3}$

Answer 1

Answer: (B)

$\frac{1}{3}$

Answer 2

When A + B = 60⁰ \ B = 60⁰ – A

\ tan B = tan (60⁰ – A) = $\frac{\sqrt{3} - \tan A}{1 + \sqrt{3}\tan A}$

Now z = tan A tan B

or z = $\frac{t(\sqrt{}3 - t)}{1 + \sqrt{}3t}$ = $\frac{\sqrt{}3t - t^{2}}{1 + \sqrt{}3t}$

where t = tan A

$\frac{dz}{dt}$ = $\frac{(t + \sqrt{}3)(\sqrt{}3t - 1)}{(1 + \sqrt{}3t)^{2}}$ = 0

\ t = 1/Ö3

\ t = tan A = tan 30⁰

\ B = 60⁰ – A = 30⁰

The other value is rejected as both A and B are +ive acute angles.

If t < $\frac{1}{\sqrt{}3}$, $\frac{dz}{dt}$ = + ive and if t > $\frac{1}{\sqrt{}3}$, $\frac{dz}{dt}$ = – ive.

Hence max. when t = $\frac{1}{\sqrt{}3}$ and max. value = $\frac{1}{3}$.