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Q)

# Prove that $\tan\alpha\tan(60^{\large\circ}-\alpha)\tan (60^{\large\circ}+\alpha)=\tan 3\alpha$

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A)
Toolbox:
• $\sin (A+B)\sin(A-B)=\sin^2A-\sin^2B$
• $\cos (A+B)\cos(A-B)=\cos^2A-\sin^2B$
• $\sin 3A=3\sin A-4\sin^3A$
• $\cos 3A=4\cos^3 A-3\cos A$
LHS
$\tan\alpha\tan(60^{\large\circ}-\alpha)\tan (60^{\large\circ}+\alpha)$
$\large\frac{\sin \alpha}{\cos\alpha}.\frac{\sin (60^{\large\circ}-\alpha)}{\cos (60^{\large\circ}-\alpha)}.\frac{\sin (60^{\large\circ}+\alpha)}{\cos (60^{\large\circ}+\alpha)}$
$\Rightarrow \large\frac{\sin \alpha(\sin^260^{\large\circ}-\sin^2\alpha)}{\cos \alpha(\cos^260^{\large\circ}-\sin^2\alpha)}$
$\Rightarrow\large\frac{ \sin\alpha(\Large\frac{3}{4}-\sin^2\alpha)}{\cos\alpha(\Large\frac{1}{4}-\sin^2\alpha)}$
$\Rightarrow\large\frac{ \sin\alpha(3-4\sin^2\alpha)}{\cos\alpha(1-4\sin^2\alpha)}$
$\Rightarrow\large\frac{ \sin\alpha(3-4\sin^2\alpha)}{\cos\alpha(1-4(1-\cos^2\alpha))}$
$\Rightarrow\large\frac{ 3\sin\alpha-4\sin^3\alpha}{\cos\alpha(4\cos^2\alpha-3)}$
$\Rightarrow\large\frac{ 3\sin\alpha-4\sin^3\alpha}{4\cos^3\alpha-3\cos \alpha}$
$\Rightarrow \large\frac{\sin 3\alpha}{\cos 3\alpha}$
$\Rightarrow \tan 3\alpha$
Hence proved