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What is the easiest way to calculate :

$$(\sqrt{3} + \tan (1^\circ)).((\sqrt{3} +\tan(2^\circ))...((\sqrt{3}+\tan(29^\circ)) $$

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This is Problem J39 from Volume 4 of Mathematical Reflections. – Byron Schmuland Aug 30 '12 at 15:21
up vote 18 down vote accepted

$\sqrt3+\tan(30^\circ-x)=\sqrt3+\frac{\frac{1}{\sqrt3}-\tan x}{1+ \frac{1}{\sqrt3}\tan x}$ $=\sqrt 3+\frac{1-\sqrt 3 \tan x}{\sqrt 3+\tan x}=\frac{4}{\sqrt 3+\tan x}$

$$\Longrightarrow(\sqrt 3+\tan(30^\circ-x))(\sqrt 3+\tan x)=4$$

Put $x=1,2,....,14^\circ$.

To complete, $x=15^\circ$, $(\sqrt 3+\tan(30^\circ-15^\circ))(\sqrt 3+\tan 15^\circ)=4$

$\implies (\sqrt 3+\tan 15^\circ)^2=4\implies \sqrt 3+\tan 15^\circ=2$ as $\tan 15^\circ>0$.

So, the answer should be $4^{14}\cdot 2=2^{29}$

A little generalization : Assuming $A≠n\frac{\pi}{2}$(where $n$ is any integer) so that $\cot A$ and $\tan A$ are non-zero finite,

$\cot A+ \tan(A-y)= \cot A+ \frac{\tan A-\tan y}{1+\tan A\tan y}=\frac{\cot A + \tan A}{1+\tan A\tan y}=\frac{\csc^2A}{\cot A + \tan y}$ (multiplying the numerator & the denominator by $\cot A$)

$\implies (\cot A+ \tan(A-y))(\cot A + \tan y)=\csc^2A $ , here $A=30^\circ$.

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+1 Very nice, concise, clear. – DonAntonio Aug 30 '12 at 9:28
How does one come up with this kind of stuff?? – NikolajK Aug 30 '12 at 9:41
@Nick Kidman: it's simple. You need some practice and the rest will come. – user 1618033 Aug 30 '12 at 10:14

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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This is great. I think it's even more clear if you do the $\cos(x)=\sin(90°-x)$ step first: You show that $\sqrt{3} + \tan(x)=2\frac{\cos(30-x)}{\cos(x)}$, and so when the product makes $x$ run from $1°$ to $29°$, all the $\cos$'s cancel. – NikolajK Aug 30 '12 at 9:54
@Nick I see what you mean, the way you suggest I think is a better way to set it out in the end, but if you haven't solved the question yet and this is the first time you're seeing a solution to it, I think this is a clearer way to see just how the sin's and cos's cancel. Though, in the end, it's just a matter of personal preference I suppose. – Monty Gill Aug 30 '12 at 14:46

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 $$\frac{P}{2^{29}}=\frac{\sin61^\circ\sin62^\circ\cdots\sin89^\circ}{\cos1^\circ\cos2^\circ\cdots\cos29^\circ}=\frac{\sin61^\circ\sin62^\circ\cdots\sin89^\circ}{\sin89^\circ\sin88^\circ\cdots\sin61^\circ}=1$$ $$P={2^{29}}.$$


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