# How to evaluate the finite power tower $\tan(1°)^{\tan(2°)^{\tan(3°)^{\cdot^{\cdot^{\cdot^{\tan(44°)^{\tan(45°)}}}}}}}$

Consider the following finite power tower:

$$\Large \tan(1°)^{\tan(2°)^{\tan(3°)^{\cdot^{\cdot^{\cdot^{\tan(44°)^{\tan(45°)}}}}}}}$$

I'm wondering if there is a way to solve this that doesn't rely on brute force (i.e. simply typing the entire power tower out). As, I see it, there are two possibilities:

1. There's some magic trig identity that simplifies this to a small closed form. I haven't found any such yet.
2. I need to develop some sort of approximation scheme for a computer.

Is (1) possible? If not, how do I do (2)?

Edit: For clarification, I have manually typed out the problem inside Desmos and arrived at the following answer:

I'd just like to know if this answer is correct, and if so, how I would go about finding an easier way to achieve this answer (whether that be by using a trig identity or some form of computerization, I don't mind) without having the type out every step like I've done above. Thank you,

• Why do people keep downvoting? What's wrong with the question? Why not provide constructive criticism instead of being actively malicious? Commented Jun 30, 2021 at 5:07
• You've given no real reason why you can't use a computer to approximate this, though it is obviously going to be very tiny. Furthermore, the question cannot be generalized, as you'll be raising to the power of infinity at 90 degrees, so I think it would be easier to write a for loop that can do this. It doesn't look like it would provide insight into other problems. Commented Jun 30, 2021 at 5:16
• You can omit the last entry since it is $1$ , but apart from that there is no shortcut. Best is to calculate the power tower iteratively. I do not expect a nice closed form. Commented Jun 30, 2021 at 5:38
• @AussieMathematician: I've edited your question to make it a little more answerable. Please check that it conforms with your intent. Commented Jun 30, 2021 at 5:53
• If it does, please edit into the post what sorts of trig identities you know for $\tan$, so we can explain why you are unlikely to see a simplification. Commented Jun 30, 2021 at 5:53

Taking them five at the time, the results are $$\left( \begin{array}{ccc} \text{from} & \text{to} & \text{value} \\ 1 & 5 & 0.0535625370657532607362393 \\ 1 & 10 & 0.4575153711252385756282372 \\ 1 & 15 & 0.2482699982569751321101037 \\ 1 & 20 & 0.2797539387187735772940906 \\ 1 & 25 & 0.2773295827116058036618656 \\ 1 & 30 & 0.2774030591794901804478588 \\ 1 & 35 & 0.2774023124430804526117091 \\ 1 & 40 & 0.2774023141041003428422647 \\ 1 & 45 & 0.2774023141038327352949884 \end{array} \right)$$
For the fun of it, with an absolute error of $$5.79\times 10^{-21}$$, the final value is almost $$\Bigg[e^{-42-\frac{34}{e}-47 e+\frac{7}{\pi }+30 \pi }\,\, \pi ^{23 e-15} \,\,\tan ^{31}(e \pi )\,\, \sec ^{13}(e \pi ) \Bigg]^{\frac 1 {15}}$$ given by a friend who enjoys this kind of problems.