Question
Verify that \tan 3\theta = 4\tan \theta\cdotp \tan(60\degree -\theta )\cdotp \tan(60\degree +\theta )
Verify that \tan 3\theta = 4\tan \theta\cdotp \tan(60\degree -\theta )\cdotp \tan(60\degree +\theta )
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,
4\tan \theta\cdotp \tan(60\degree -\theta )\cdotp \tan(60\degree +\theta ) =\tan 3\theta