<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] $$\tan^2C=\cos^2B=\dfrac1{1+\tan^2B}=\dfrac1{1+\cos^2A}=\dfrac1{2-\sin^2A}$$
$$\cos^2C=\tan^2A$$
$$\implies\tan^2C=\sec^2C-1=\dfrac{1-\cos^2C}{\cos^2C}=\dfrac{1-\tan^2A}{\tan^2A}=\dfrac{1-2\sin^2A}{\sin^2A}$$
Equating the values of $\tan^2C$ and writing $\sin^2A=x$
$$\dfrac1{2-x}=\dfrac{1-2x}x\implies x^2-3x+1=0\implies x=\dfrac{3-\sqrt5}2$$
As $0\le x\le1$
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