Prove that tan x tan
Question:

Prove that $\tan x \tan \left(\frac{\pi}{3}-x\right) \tan \left(\frac{\pi}{3}+x\right)=\tan 3 x$

Solution:

$\frac{\pi}{3}=60^{\circ}$

LHS $=\tan x \tan \left(60^{\circ}-x\right) \tan \left(60^{\circ}+x\right)$

$=\frac{\sin x \sin \left(60^{\circ}-x\right) \sin \left(60^{\circ}+x\right)}{\cos x \cos \left(60^{\circ}-x\right) \cos \left(60^{\circ}+x\right)}$

$=\frac{\sin x\left(\sin ^{2} 60^{\circ}-\sin ^{2} x\right)}{\cos x\left(\cos ^{2} 60^{\circ}-\sin ^{2} x\right)}$

$=\frac{\sin x\left(\frac{3}{4}-\sin ^{2} x\right)}{\cos x\left(\frac{1}{4}-\sin ^{2} x\right)}$

$=\frac{\sin x\left(3-4 \sin ^{2} x\right)}{\cos x\left(1-4 \sin ^{2} x\right)}$

$=\frac{3 \sin x-4 \sin ^{3} x}{4 \cos ^{3} x-3 \cos x}$

$=\frac{\sin 3 x}{\cos 3 x}$

$=\tan 3 x=\mathrm{RHS}$

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