The minimum value of  ${[{(\tan A-\cos B)}^2+{(4\cot A-\sin B)}^2]}^{1/2}$, where A and B are independent defined variables, is of the form  $b\sqrt{c}-a$, where a, b, c are natural numbers and c is prime. The value of $a + b + c$ is equal to

The given expression can be interpreted as the square of the distance between the points $(\tan A, 4cotA) and (\cos B, SinB).$
The minimum value of this distance is the minimum distance between the curves 
$xy = 4$ and  $x^2+y^2=1$
$P(t,\frac{4}{t}),Q(0,0)$

Least distance =  $PQ-1=(\sqrt{t^2+\frac{16}{t^2}}-1)\ge \sqrt{{(t-\frac{4}{t})}^2+8}-1=\sqrt{8}-1=2\sqrt{2}-1$

$\ge 4\sqrt{2}-1$