2026/045) Prove that $(4\cos^2 9^{\circ}–3)(4\cos^2 27^{\circ}–3)=\tan 9^{\circ}$

We shall use formula for $\cos 3x$  

$\cos 3x = 4 \cos^3 x - 3 \cos x$

so we have $4 \cos ^2 x - 3 = \frac{\cos 3x}{ \cos x}$

Hence 

$4 \cos ^2 9^{\circ} - 3 = \frac{\cos 27^{\circ}}{\cos 9^{\circ}}$

And 

$4 \cos ^2 27^{\circ} - 3 = \frac{\cos 81^{\circ}}{\cos 27^{\circ}}$

so   $(4\cos^2 9^{\circ}–3)(4\cos^2 27^{\circ}–3)$ = $\frac{\cos 27^{\circ}}{\cos 9^{\circ}}$ *  $\frac{\cos 81^{\circ}}{\cos 27^{\circ}}$ 

= $\frac{\cos 81^{\circ}}{\cos 9^{\circ}}$

 = $\frac{\sin 9^{\circ}}{\cos 9^{\circ}}$ using $\cos \theta = \sin (90^{\circ}-\theta)$

 = $\tan 9^{\circ}$

Hence Proved