TITLE: Calculating :$((\sqrt{3} + \tan (1^\circ)).((\sqrt{3} +\tan(2^\circ))...((\sqrt{3}+\tan(29^\circ))$
QUESTION [11 upvotes]: What is the easiest way to calculate :
$$(\sqrt{3} + \tan (1^\circ)).((\sqrt{3} +\tan(2^\circ))...((\sqrt{3}+\tan(29^\circ)) $$

REPLY [4 votes]: Maybe you wanna go this way:
$$P=(\sqrt{3} + \tan (1^\circ))((\sqrt{3} +\tan(2^\circ))\cdots((\sqrt{3}+\tan(29^\circ))$$
$$\frac{P}{2^{29}}=\frac{(\sqrt{3} \cos(1^\circ) + \sin (1^\circ))}{2\cos(1^\circ)}\frac{(\sqrt{3} \cos(2^\circ) + \sin (2^\circ))}{2\cos(2^\circ)}\cdots\frac{(\sqrt{3} \cos(29^\circ) + \sin (29^\circ))}{2\cos(29^\circ)}$$
Above I used the fact that
$$\sin(60^\circ+\alpha^\circ)=\frac{\sqrt3}{2}\cos \alpha^\circ+\frac{1}{2}\sin \alpha^\circ$$
then
$$\frac{P}{2^{29}}=\frac{\sin61^\circ\sin62^\circ\cdots\sin89^\circ}{\cos1^\circ\cos2^\circ\cdots\cos29^\circ}=\frac{\sin61^\circ\sin62^\circ\cdots\sin89^\circ}{\sin89^\circ\sin88^\circ\cdots\sin61^\circ}=1$$
$$P={2^{29}}.$$
Q.E.D.