$\tan\theta tan\left(\frac{\pi }{3}+\theta \right)\tan \left(\frac{\pi }{3}-\theta \right)$=$tan3\theta$
Prove that tanθ tan(π/3+θ) tan(π/3-θ)=tan3θ
Solution:
$\tan \theta tan\left(\frac{\pi }{3} + \theta \right)\tan \left(\frac{\pi }{3} – \theta \right)$
= $\frac{\sin \theta }{\cos \theta } \times \frac{\sin \left(\frac{\pi }{3} + \theta \right)sin\left(\frac{\pi }{3} – \theta \right)}{\cos \left(\frac{\pi }{3} + \theta \right)cos\left(\frac{\pi }{3} – \theta \right)}$
= $\frac{\sin \theta }{\cos \theta } \times \frac{2\sin \left(\frac{\pi }{3} + \theta \right)sin\left(\frac{\pi }{3} – \theta \right)}{2\cos \left(\frac{\pi }{3} + \theta \right)cos\left(\frac{\pi }{3} – \theta \right)}$
= $= \frac{\sin \theta }{\cos \theta } \times \frac{\cos { (\frac{\pi }{3} + \theta ) – (\frac{\pi }{3} – \theta )} – cos{ (\frac{\pi }{3} + \theta ) + (\frac{\pi }{3} – \theta )} }{\cos { (\frac{\pi }{3} + \theta ) + (\frac{\pi }{3} – \theta )} + cos{ (\frac{\pi }{3} + \theta ) – (\frac{\pi }{3} – \theta )} }$
= $\frac{\sin \theta }{\cos \theta } \times \frac{\cos 2\theta – cos\frac{2\pi }{3}}{\cos \frac{2\pi }{3} + cos2\theta }$
= $\frac{\sin \theta }{\cos \theta } \times \frac{\cos 2\theta – cos\left(\pi – \frac{\pi }{3}\right)}{\cos \left(\pi – \frac{\pi }{3}\right) + cos2\theta }$
= $\frac{\sin \theta }{\cos \theta } \times \frac{\cos 2\theta + cos\frac{\pi }{3}}{ – \cos \frac{\pi }{3} + cos2\theta }$
= $\frac{\sin \theta }{\cos \theta } \times \frac{\cos 2\theta + \frac{1}{2}}{ – \frac{1}{2} + \cos 2\theta }$
= $\frac{\sin \theta }{\cos \theta } \times \frac{2cos2\theta + 1}{ – 1 + 2cos2\theta } $
= $\frac{2cos2\theta sin\theta + sin\theta }{2cos2\theta cos\theta – cos\theta }$
= $\frac{\sin \left(2\theta + \theta \right) – sin\left(2\theta – \theta \right) + sin\theta }{\cos (2\theta + \theta ) + cos\left(2\theta – \theta \right) – cos\theta }$
= $\frac{\sin 3\theta – sin\theta + sin\theta }{\cos 3\theta + cos\theta – cos\theta }$
= $tan3\theta$ [Proved]