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.

If no two of $x,y,z$ are equal then all of them are unequal. Assume $x>y>z.$ From $(1) and (2)$ $y - z + y(x - z) = 0 $. This implies $ y < 0 $(since $y - z > 0, x - z > 0)$.....$(4)$. From $(2) and (3)$:$ z -x + z(y - x) = 0 $
This implies $z < 0$ (since $z - x < 0, y - x < 0)$ $......(5)$. 
From $(3) and (1)$: $x - y + x(z - y) = 0 $
This implies $x > 0$(since $x - y > 0, z - y < 0$)$.........(6)$ 
(4) and (6) contradicts (1), which has (1 + x)y = 1 
→ our assumption that $x, y, z$ are all unequal is incorrect and 
some two of them are equal. If $x = y$, $(2)$ becomes 
$(1 + x)z = 1 = (1 + z)x$ (by $(3)$) 
→$1 - z = xz = 1 - x → z = x→ x = y = z$. 
If $y = z$, it similarly follows that $x = y = z$. 
So we have: $tan²A = tan²B = tan²C$ 
→$A = B = C →sinA = sinB = sinC$.
