Product of Sum of Tangents?

Let $f(n) = \sqrt3 + \tan(n^\circ)$ .

We know that $f(n)$ can be written by:

$f(n) = \dfrac{1}{\tan(30^\circ)} + \tan(n^\circ) = \dfrac{\cos(30^\circ)}{\sin(30^\circ)} + \frac{\sin(n^\circ)}{\cos(n^\circ)} = \dfrac{\cos(n^\circ)\cos(30^\circ) + \sin(n^\circ)\sin(30^\circ)}{\cos(n^\circ)\sin(30^\circ)} = \dfrac{\cos(30^\circ - n^\circ)}{\cos(n^\circ)} \dfrac{1}{\sin(30^\circ)}$ .

Therefore, $\displaystyle \prod^{29}_{n=1} f(n) = \dfrac{1}{\sin^{29}(30^\circ)} \dfrac{\cos(29^\circ)\cos(28^\circ)\cos(27^\circ)...\cos(1^\circ)}{\cos(1^\circ)\cos(2^\circ)\cos(3^\circ)...\cos(29^\circ)} = 2^{29}.$

Thus, $x = \boxed{29}$

The above expression can be : 2sin(60+x)/cos(x) put consecutively the values and you get x=29.

let α=(3√+tan(1)).(3√+tan(2)).(3√+tan(3))......(3√+tan(29)) Which can be written as α=(3√+tan(1)).(3√+tan(29)).(3√+tan(2)).(3√+tan(28)).......(3√+tan(14)).(3√+tan(16)).(3√+tan(15))

Now,, notice that (3√+tan(1)).(3√+tan(29))=3+3√(tan(1)+tan(29))+tan(1).tan(29)...........(1)

Now we know from the formula that tan(30)=tan(1)+tan(29)1−tan(1)tan(29) Therefore, tan(1)+tan(29)=13√(1−tan(1)tan(29)) , so putting this value in (1), we get (3√+tan(1)).(3√+tan(29))=3+(1−tan(1).tan(29))+tan(1).tan(29)=4

So, like this, all the pairs cancel out to give 4 in each bracket, So, the original problem becomes α=414((3√+tan(15))

Now, we see that (3√+tan(15))=3√+sin(30)1+cos(30)=2

So, the original number α is nothing but, α=(414).2