Question

Prove that $\tan x\tan \left(\frac{\pi }{3}-x\right)\tan \left(\frac{\pi }{3}+x\right)=\tan 3x$

Answer 1

$\frac{\pi }{3}=60°$
$$\begin{gathered} \mathrm{LHS}=\tan x\tan \left(60°-x\right)\tan \left(60°+x\right) \\ =\frac{\sin x\sin \left(60°-x\right)\sin \left(60°+x\right)}{\cos x\cos \left(60°-x\right)\cos \left(60°+x\right)} \end{gathered}$$

$$\begin{gathered} =\frac{\sin x\left({\sin }^{2}60°-{\sin }^{2}x\right)}{\cos x\left({\cos }^{2}60°-{\sin }^{2}x\right)} \\ =\frac{\sin x\left({\displaystyle \frac{3}{4}}-{\sin }^{2}x\right)}{\cos x\left({\displaystyle \frac{1}{4}}-{\sin }^{2}x\right)} \\ =\frac{\sin x\left({\displaystyle 3}-4{\sin }^{2}x\right)}{\cos x\left({\displaystyle 1}-4{\sin }^{2}x\right)} \end{gathered}$$

$$\begin{gathered} =\frac{3\sin x-4{\sin }^{3}x}{4{\cos }^{3}x-3\cos x} \\ =\frac{\sin 3x}{\cos 3x} \\ =\tan 3x=\mathrm{RHS} \end{gathered}$$