<s>[INST] Calculating :$((\sqrt{3} + \tan (1^\circ)).((\sqrt{3} +\tan(2^\circ))...((\sqrt{3}+\tan(29^\circ))$ What is the easiest way to calculate :
$$(\sqrt{3} + \tan (1^\circ)).((\sqrt{3} +\tan(2^\circ))...((\sqrt{3}+\tan(29^\circ)) $$
 [/INST] Note that $\sqrt{3} = \tan(60^{\circ})$, so your product is $$(\tan(60^{\circ}) + \tan (1^\circ))\cdot((\tan(60^{\circ}) +\tan(2^\circ))\cdots((\tan(60^{\circ})+\tan(29^\circ)).$$ Now use the identity (which is very simple to prove by just writing $\tan$ in terms of $\sin$ and $\cos$ and adding the resulting fractions) $$\tan(A) + \tan(B) = \frac{\sin(A+B)}{\cos(A)\cos(B)}.$$ For your product, set $A = 60^{\circ}$, so we have $$(\tan(60^{\circ}) + \tan (1^\circ))\cdot((\tan(60^{\circ}) +\tan(2^\circ))\cdots((\tan(60^{\circ})+\tan(29^\circ)) \\ = \frac{\sin(61^{\circ})\sin(62^{\circ})\cdots\sin(89^{\circ})}{(\cos(60^{\circ}))^{29}\cos(1^{\circ})\cos(2^{\circ})\cdots\cos(29^{\circ})}.$$ And using $\cos(x)=\sin(90^{\circ}-x)$, this reduces to $$\frac{1}{(\cos(60^{\circ}))^{29}}=2^{29}.$$
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