A "fast" way for computing $ \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

1. reindexed the product in the numerator, and
2. 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.

Using this, $$(\cot A + \tan y)(\cot A+ \tan(A-y))=\csc^2A \text{ if } A\ne m\pi\text{ where }m\text{ is any integer}$$

Putting $A=45^\circ, (1 + \tan y)(1+ \tan(45^\circ-y))=\csc^245^\circ=2$

Now, putting $y=1^\circ,2^\circ,3^\circ,\cdots,\lfloor\frac{45}2\rfloor^\circ=22^\circ$ and multiplying them we get, $$\prod_{1\le r\le 22}(1+\tan r^\circ)(1+\tan(45-r)^\circ)=2^{22}$$

$$\implies \prod_{1\le r\le 44}(1+\tan r^\circ)=2^{22}$$

The unpaired $1+\tan45^\circ=1+1=2$