Question

Prove that tan θ tan (60° − θ) tan (60° + θ) = tan 3θ.

Answer 1

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

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

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