<s>[INST] If $\cos A=\tan B$, $\cos B=\tan C$ … If $\cos A=\tan B$, $\cos B=\tan C$ and $\cos C=\tan A$, prove that $\sin A=\sin B=\sin C$.
My Attempt.
Let us consider $x$, $y$ and $z$ as:.
$$x = \tan^2A$$
$$y = \tan^2B$$
$$z = \tan^2C$$
$$\cos^2A = \tan^2B$$
$$\frac {1}{\sec^2A}= \tan^2B$$
$$\frac {1}{1 + \tan^2A} = \tan^2B$$
$$\frac {1}{1 + x} = y$$
$$(1 + x)y = 1\tag{1}$$
Similarly,
$$(1 + y)z = 1\tag{2}$$
$$(1 + z)x = 1\tag{3}$$
Please help me to continue from here.
 [/INST] Define $a := \cos^2 A$, and $b := \cos^2 B$, and $c := \cos^2 C$. Then 
$$\cos A = \tan B \quad\to\quad \cos^2 A = \tan^2 B = \sec^2 B - 1 \quad\to\quad a = \frac{1}{b} - 1 \quad\to\quad b = \frac{1}{1+a}$$
Likewise,
$$c = \frac{1}{1+b} \qquad\text{and}\qquad a = \frac{1}{1+c}$$
so that, adding a slightly-gratuitous $1$, 
$$1+a \;=\; 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+a}}} \\ \tag{$\star$}$$
The evident recursion reveals that $1+a$ (and also $1+b$ and $1+c$) is the "$1$s all the way down" continued fraction, which some will recognize as representing the Golden Ratio, $\phi := \frac{1}{2}(1+\sqrt{5}) = 1.618\ldots$. Consequently,
$$a = b = c = \phi - 1 = \phi^{-1}$$
We finally calculate
$$\sin^2 A = \sin^2 B = \sin^2 C = 1 - \phi^{-1} = \phi^{-2}$$
so that 

$$|\sin A| = |\sin B| = |\sin C| = \phi^{-1} = 0.618\ldots$$

That the signs of the sines match is left as an exercise to the reader.
Observe that we'd reach the same conclusion no matter the length of the  "loop" of equations. Once $1+a$ appears, at any stage, in the right-hand-side of the counterpart of $(\star)$, the implied continued fraction collapses to the simpler form 
$$1 + a = 1 + \frac{1}{1+a} \quad\text{, giving}\quad a = \frac{1}{1+a}$$
which represents the one-equation loop, $\cos A = \tan A$.
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