The Value of \( \tan x \tan\left(\frac{\pi}{3}-x\right)\tan\left(\frac{\pi}{3}+x\right) \)
Question
Find the value of
\[ \tan x\, \tan\left(\frac{\pi}{3}-x\right)\, \tan\left(\frac{\pi}{3}+x\right) \]
(a) \(\cot3x\)
(b) \(2\cot3x\)
(c) \(\tan3x\)
(d) \(3\tan3x\)
Solution
Let
\[ t=\tan x \]
Using
\[ \tan\left(\frac{\pi}{3}-x\right) = \frac{\sqrt3-t}{1+\sqrt3\,t} \]
and
\[ \tan\left(\frac{\pi}{3}+x\right) = \frac{\sqrt3+t}{1-\sqrt3\,t} \]
Therefore,
\[ \tan\left(\frac{\pi}{3}-x\right) \tan\left(\frac{\pi}{3}+x\right) = \frac{(\sqrt3-t)(\sqrt3+t)} {(1+\sqrt3 t)(1-\sqrt3 t)} \]
\[ = \frac{3-t^2} {1-3t^2} \]
Multiplying by \(\tan x=t\),
\[ \text{LHS} = \frac{t(3-t^2)} {1-3t^2} \]
Now use the identity
\[ \tan3x = \frac{3t-t^3} {1-3t^2} = \frac{t(3-t^2)} {1-3t^2} \]
Hence,
\[ \tan x \tan\left(\frac{\pi}{3}-x\right) \tan\left(\frac{\pi}{3}+x\right) = \tan3x \]
Final Answer
\[ \boxed{\tan3x} \]
Hence, the correct option is (c) \(\tan3x\).