A "fast" way for  computing $ \prod \limits_{i=1}^{45}(1+\tan i^\circ) $? 
Which  is the fastest paper-pencil approach to compute the product $$ \prod
 \limits_{i=1}^{45}(1+\tan i^\circ) $$


Using
$$
1+\tan x = \frac{\sin x + \cos x}{\cos x} = \frac{\sqrt{2} \cos (45^{\circ} - x)}{\cos x}, 
$$
the product can be written as:
$$
\prod_{x=1}^{45}(1+\tan x^\circ) = 2^{45/2} \prod_{x=1}^{45} \frac{\cos (45 - x)^{\circ}}{\cos x^{\circ}} \stackrel{(1)}{=}  2^{45/2}  \cdot \frac{\prod\limits_{x=0}^{44} \cos x^{\circ}}{\prod\limits_{x=1}^{45} \cos x^{\circ}} \stackrel{(2)}{=} 2^{45/2} \cdot \frac{\cos 0}{\cos 45^{\circ}} = 2^{23},
$$
where we


*

*reindexed the product in the numerator, and

*cancelled the common factors. 



Another approach. If $x+y = 45^{\circ}$, then
$$
1 = \tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y},
$$
which rearranges to
$$
\tan x \tan y + \tan x + \tan y = 1 \quad \implies \quad (1+\tan x)(1+\tan y) = 2.
$$
Now plug in $x = 0^{\circ}, 1^{\circ}, 2^{\circ}, \ldots, 45^{\circ}$, so that $y$ takes the same values but in the opposite order. Multiplying all these equations, we get
$$
\left[ \prod_{x=0}^{45} (1+\tan x^\circ) \right]^2 = 2^{46}.
$$
Taking square-roots and noting that $1+\tan 0^\circ = 1$, we get the answer. 
