4\tan \theta\cdotp \tan(60\degree -\theta )\cdotp \tan(60\degree +\theta )

=4\cdotp \dfrac{\sin \theta\cdotp \sin(60\degree -\theta )\cdotp \sin(60\degree +\theta ) }{\cos \theta\cdotp \cos(60\degree -\theta )\cdotp \cos(60\degree +\theta ) }

4\sin \theta \cdotp \sin(60\degree -\theta )\cdotp \sin(60+\theta )

=2\sin \theta (\cos2\theta -\cos 120\degree )

=2\sin \theta (\cos2\theta+\dfrac{1}{2} )

=2\sin \theta (\dfrac{3}{2}-2\sin^2 \theta )

=3\sin \theta -4\sin^3 \theta

=\sin 3\theta

4\cos \theta \cdotp \cos(60\degree -\theta )\cos(60\degree +\theta )

=2\cos \theta (\cos 120\degree +\cos 2\theta )

=2\cos \theta (-\dfrac{1}{2}+2\cos^2 \theta -1)

=-3\cos \theta +4\cos^3 \theta

=\cos 3\theta

Therefore,

\tan \theta\cdotp \tan(60\degree -\theta )\cdotp \tan(60\degree +\theta ) =\tan 3\theta

Using triple angle identity for tangent

\tanθ\tan(60\degree−\tanθ )\tan(60\degree +θ)

=\tanθ⋅\dfrac{\tan60\degree- \tanθ}{1+\tan60\degree\tanθ }\cdot \dfrac{\tan60\degree+\tanθ }{1-\tan60\degree\tanθ}

=\tanθ⋅\dfrac{\sqrt{3} - \tanθ}{1+\sqrt{3} \degree\tanθ }\cdot \dfrac{\sqrt{3} +\tanθ }{1-\sqrt{3} \degree\tanθ}

=\dfrac{3\tanθ -\tan^3θ }{1-3\tan^2θ }

=\tan3θ