Question
Prove that:
$\tan\text{x}\tan\Big(\frac{\pi}{3}-\text{x}\Big)\tan\Big(\frac{\pi}{3}+\text{x}\Big)=\tan3\text{x}$

Answer

$\frac{\pi}{3}=60^\circ$
$\text{LHS}=\tan\text{x}\tan(60^\circ-\text{x})\tan(60^\circ+\text{x})$
$=\ \frac{\sin\text{x}\sin(60^\circ-\text{x})\sin(60^\circ+\text{x})}{\cos\text{x}\cos(60^\circ-\text{x})\cos(60^\circ+\text{x})}$
$=\ \frac{\sin\text{x}(\sin^260^\circ-\sin^2\text{x})}{\cos\text{x}(\cos^260^\circ-\sin^2\text{x})}$
$=\ \frac{\sin\text{x}\Big(\frac{3}{4}-\sin^2\text{x}\Big)}{\cos\text{x}\Big(\frac{1}{4}-\sin^2\text{x}\Big)}$
$=\ \frac{\sin\text{x}(3-4\sin^2\text{x})}{\cos\text{x}(1-4\sin^2\text{x})}$
$=\ \frac{3\sin\text{x}-4\sin^3\text{x}}{4\cos^3\text{x}-3\cos\text{x}}$
$=\ \frac{\sin3\text{x}}{\cos3\text{x}}$
$=\ \tan3\text{x}=\text{RHS}$

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