Why is $(1-\cot 37^\circ)(1-\cot 8^\circ)=2.00000000\cdots$? Apparently, 
$$(1-\cot 37^\circ)(1-\cot 8^\circ)=2.00000000000000000\cdots$$
Since it is a $2.0000000000\cdots$ instead of $2$, it isn't exactly $2$.
Why is that?
 A: Using various trigonometric identities, you should be able to show the left hand side is equal to $2$. The ones you will need are:
$$\cos(A+B)=\cos A\cos B-\sin A\sin B,\quad \cos(-A)=\cos A,\quad \cos(45^\circ)=\sin(45^\circ)=\frac1{\sqrt{2}}$$
as well as an understanding of what $\cot A$ means. The reason you have the answer of $2.000000\ldots$ (which I assume you got from WolfRamAlpha or similar) is because it's easier for a computer to just work out $\cot 37^\circ,\cot 8^\circ$ to any degree of accuracy and then solve by brute force. So the computer gets an answer of, for example, $2.000000$ correct to $6$ decimal places, but cannot be sure that the answer is exactly $2$.
A: $$\begin{align}
1-\cot{\left(x\right)}
&=1-\frac{\cos{\left(x\right)}}{\sin{\left(x\right)}}\\
&=\frac{\sin{\left(x\right)}-\cos{\left(x\right)}}{\sin{\left(x\right)}}\\
&=-\sqrt{2}\frac{\sin{\left(\frac{\pi}{4}-x\right)}}{\sin{\left(x\right)}}.\\
\end{align}$$
Therefore,
$$\begin{align}
1-\cot{\left(\frac{\pi}{4}-x\right)}
&=-\sqrt{2}\frac{\sin{\left(\frac{\pi}{4}-\left(\frac{\pi}{4}-x\right)\right)}}{\sin{\left(\frac{\pi}{4}-x\right)}}\\
&=-\sqrt{2}\frac{\sin{\left(x\right)}}{\sin{\left(\frac{\pi}{4}-x\right)}}.\\
\end{align}$$
Hence,
$$\begin{align}
\left(1-\cot{\left(x\right)}\right)\cdot\left(1-\cot{\left(\frac{\pi}{4}-x\right)}\right)
&=\left[-\sqrt{2}\frac{\sin{\left(\frac{\pi}{4}-x\right)}}{\sin{\left(x\right)}}\right]\cdot\left[-\sqrt{2}\frac{\sin{\left(x\right)}}{\sin{\left(\frac{\pi}{4}-x\right)}}\right]\\
&=2.\
\end{align}$$
A: $$
1 = \cot45^\circ = \cot(37^\circ+8^\circ) = \frac{\cot37^\circ\cot8^\circ-1}{\cot37^\circ+\cot8^\circ}.
$$
Therefore
$$
\cot37^\circ+\cot8^\circ = \cot37^\circ\cot8^\circ-1
$$
so
$$
2 = 1-\cot37^\circ-\cot8^\circ +\cot37^\circ\cot8^\circ=(1-\cot37^\circ)(1-\cot8^\circ).
$$
A: Alternatively,
$$\cot(x+y) = \frac{\cos(x+y)}{\sin(x+y)} = \frac{\cos x \cos y - \sin x \sin y}{\sin x \cos y + \cos x \sin y} = \frac{\cot x \cot y - 1}{\cot x + \cot y}$$
Now if $x = 37^\circ$, $y = 8^\circ$, then $\cot(x + y) = \cot 45^\circ = 1$ and
$$\begin{align}  \ \ \ \  1 & = \frac{\cot x \cot y - 1}{\cot x + \cot y} \\  \cot x + \cot y & = \cot x \cot y - 1 \\
   2 + \cot x + \cot y & = \cot x \cot y + 1 \\
 1 - \cot x - \cot y + \cot x \cot y  & = 2 \\
(1 - \cot x)(1 - \cot y) & = 2 \\
(1 - \cot 37^\circ)(1 - \cot 8^\circ) & = 2
\end{align}$$
A: $2.\dot 0$ is exacyly equal to 2
