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