If $\cos A=\tan B,\cos B=\tan C,\cos C=\tan A,$ then $\sin A$ is equal to

Let $\sin A=x$ ………………. ...(i)

Then, $\cos A=\tan B$

$\begin{array}{l}\Rightarrow \sqrt{1-{\sin }^2}A=\tan B\\ \Rightarrow \sqrt{1-x^2}=\tan B\end{array}$

and,$\cos B = \tan C$

$\begin{array}{l}\Rightarrow \frac{1}{\sqrt{1+{\tan }^{2}B}}=\tan C\\ \Rightarrow \frac{1}{\sqrt{2-x^2}}=\tan C\\ \Rightarrow \cos C=\frac{1}{\sqrt{1+{\tan }^{2}C}}=\sqrt{\frac{2-x^2}{3-x^2}} ...................... \left(ii\right)\end{array}$                                                 

Now, 

$\begin{array}{l}\cos C=\tan A\\ \Rightarrow \sqrt{\frac{2-x^2}{3-x^2}}=\frac{x}{\sqrt{1-x^2}}\end{array}$                     [From (i) and (ii)]

$\Rightarrow x^2=\frac{(1\pm \sqrt{5}{)}^2}{4}\Rightarrow x=\frac{\sqrt{5}-1}{2}=2\sin {18}^{\circ }$