In this blog post, we will continue to manipulate the number one to looking for the most efficient and effective solution.

According to Wikipedia (Number One):

One, sometimes referred to as unity, is the integer before two and after zero. One is the first non-zero number in the natural numbers as well as the first odd number in the natural numbers.

Any number multiplied by one is that number, as one is the identity for multiplication. As a result, one is its own factorial, its own square, its own cube, and so on. One is also the result of the empty product, as any number multiplied by one is itself. It is also the only natural number that is neither composite nor prime with respect to division, but instead considered a unit.

Mathematically, $1$ is:

in arithmetic (algebra) and calculus, the natural number that follows $0$ and precedes $2$ and the multiplicative identity element of the integers, real numbers and complex numbers;

more generally, in abstract algebra, the multiplicative identity ("unity"), usually of a ring.

One cannot be used as the base of a positional numeral system. (Sometimes tallying is referred to as "base $1$", since only one mark — the tally itself — is needed, but this is not a positional notation.)

Since the base $1$ exponential function ($1$x) always equals $1$, its inverse does not exist (which would be called the logarithm base $1$ if it did exist).

Shifting the focus to you... what do you think about $1$? Will you willing to believe that $1$ is such a powerful figure that some of the time, it helps us to solve for many not-so-obvious telescoping series problems?

Let's first observe the sum between [MATH]\color{yellow}\bbox[5px,purple]{1}\color{black}\,\, \text{and}\,\, \color{yellow}\bbox[5px,purple]{\tan n^{\circ}}[/MATH]:

$\begin{align*}1+\tan n^{\circ}&=1+\dfrac{\sin n^{\circ}}{\cos n^{\circ}}\\&=\dfrac{\cos n^{\circ}+\sin n^{\circ}}{\cos n^{\circ}}\end{align*}$

Next, let's do the sum between [MATH]\color{black}\bbox[5px,orange]{1}\color{black}\,\, \text{and}\,\, \color{black}\bbox[5px,orange]{\tan(45^{\circ}-n^{\circ})}[/MATH]:

$\begin{align*}1+\tan (45^{\circ}-n^{\circ})&=1+\dfrac{\sin (45^{\circ}-n^{\circ})}{\cos (45^{\circ}-n^{\circ})}\\&=1+\dfrac{\sin 45^{\circ}\cos n^{\circ}-\cos 45^{\circ}\sin n^{\circ}}{\cos 45^{\circ}\cos n^{\circ}+\sin 45^{\circ}\sin n^{\circ}}\\&=1+\dfrac{\sin 45^{\circ}\cos n^{\circ}-\sin 45^{\circ}\sin n^{\circ}}{\sin45^{\circ}\cos n^{\circ}+\sin 45^{\circ}\sin n^{\circ}}\\&=1+\dfrac{\cos n^{\circ}-\sin n^{\circ}}{\cos n^{\circ}+\sin n^{\circ}}\\&=\dfrac{\cos n^{\circ}+\sin n^{\circ}+\cos n^{\circ}-\sin n^{\circ}}{\cos n^{\circ}+\sin n^{\circ}}\\&=\dfrac{2\cos n^{\circ}}{\cos n^{\circ}+\sin n^{\circ}}\end{align*}$

Now, what if we multiply both sum, what would we get?

$\begin{align*}(1+\tan n^{\circ})(1+\tan (45^{\circ}-n^{\circ}))&=\dfrac{\cos n^{\circ}+\sin n^{\circ}}{\cos n^{\circ}}\cdot \dfrac{2\cos n^{\circ}}{\cos n^{\circ}+\sin n^{\circ}}\\&=\dfrac{\cancel{\cos n^{\circ}+\sin n^{\circ}}}{\cancel{\cos n^{\circ}}}\cdot \dfrac{2\cancel{\cos n^{\circ}}}{\cancel{\cos n^{\circ}+\sin n^{\circ}}}\\&=2\end{align*}$

That means, if we have

$(1+\tan 1^{\circ})(1+\tan 2^{\circ})(1+\tan 3^{\circ})(1+\tan 42^{\circ})(1+\tan 43^{\circ})(1+\tan 44^{\circ})$

It can be rewritten as

$(1+\tan 1^{\circ})(1+\tan 44^{\circ})(1+\tan 2^{\circ})(1+\tan 43^{\circ})(1+\tan 3^{\circ})(1+\tan 42^{\circ})$

It actually is equivalent to $2^3$.

So if we're asked to evaluate

$(1+\tan 1^{\circ})(1+\tan 2^{\circ})\cdots(1+\tan 43^{\circ})(1+\tan 44^{\circ})(1+\tan 45^{\circ})$

We're then be expected to (if you want to use the identity we have found above) regroup the term so we have

$(1+\tan 1^{\circ})(1+\tan 2^{\circ})\cdots(1+\tan 43^{\circ})(1+\tan 44^{\circ})(1+\tan 45^{\circ})$

$\small=\underbrace{(1+\tan 1^{\circ})(1+\tan 44^{\circ})}_{\text{first pair}}\underbrace{(1+\tan 2^{\circ})(1+\tan 43^{\circ})}_{\text{second pair}}\cdots\underbrace{(1+\tan 22^{\circ})(1+\tan 23^{\circ})}_{\text{22th pair}}(1+\tan 45^{\circ})$

$=2^{22}(1+1)$

$=2^{23}$

If I didn't believe it before reading this blog post about how the figure one can be a giant helper in solving intriguing math problems, I certainly you must have believed it now...and the list of the advantages we could get from figure $1$ is, practically speaking, endless...I will show you in another blog post the other trick we could employ of $1$ that helps us to tackle the very classic Olympiad mathematical problem so very neatly and ingeniously...

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