Given that cosine is just the sine of the complementary angle,
$$\frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{\sin(C)}{c} \tag{1}$$ so $$\frac{\cos(90^\circ-A)}{a}=\frac{\cos(90^\circ-B)}{b}=\frac{\cos(90^\circ-C)}{c} \tag{2}$$
Then, if those ratios are equal, don't we also have the following?
$$\frac{\cos(A)}{a}=\frac{\cos(B)}{b}=\frac{\cos(C)}{c} \tag{3}$$
No, try if $C=90^{\circ}$. Then ${\cos C\over c}=0$ but $\cos A\ne 0$ and $\cos B \ne 0$.
Only situation when it works it's when $a=b=c$:
If we mark $$\frac{\cos(A)}{a}=\frac{\cos(B)}{b}=\frac{\cos(C)}{c} ={1\over q}$$
then we get $a = q \cos A$, $b= q\cos B$ and $c=q\cos C$ for some positive $q$
Also from sine theorem we have $a = d \sin A$, $b= d\sin B$ and $c=d\sin C$, where $d=2R$ diameter. So we have $$1=\cos ^2A+\sin^2 A = {a^2\over q^2}+{a^2\over d^2}$$ so $${1\over a^2} = {1\over q^2}+{1\over d^2}$$ but the same holds for $b$ and $c$, so $a=b=c$ is the only case when it works.