Simplify them first

Firstly, note that $\tan A = \dfrac{\sin A}{\cos A}$ , so $\cos A \neq 0$ (since $\tan A$ would have an undefined value, so the product would have an indeterminate value). Thus, if we multiply the three real numbers, we get $\sin^2 A = 0$ , so $A=k\pi$ . If $\sin A = 0$ , then $\tan A = 0$ , and $\cos A = \pm 1$ , so the absolute value of the sum is $\boxed{1}$ .

Say $A=0$ Then sin A=0 so the product of the three real numbers is 0. So $sin A+cos A+tan A=0+1+0=1$

$tan$ $A = \frac {sin A}{cos A}$

$cos$ $A \neq 0$ , so $sin$ $A$ must be $0.$ When $sin$ $A = 0,$ $tan$ $A = 0,$ so the product of these three numbers is $:$

$sin$ $A \times cos$ $A \times tan$ $A =$

$0 \times cos$ $A \times 0 = 0$

For what values of $A$ is $sin$ $A = 0$ ? $A = 0, A = \pi$

Plug them into $cos$ $A:$

$cos$ $0 = 1$

$cos$ $\pi = -1$

Because $sin$ $A$ and $tan$ $A$ are $0,$ the sum of these three numbers is $:$

$sin$ $A + cos$ $A + tan$ $A =$

$0 + 1 + 0 = 1$

$or$

$0 - 1 + 0 = -1$

Thus, the absolute value of their sum is $1.$

Quite Easy! The product of the triplets $(\sin A,\cos A,\tan A)$ =0, if and only one term of the product is $0$ ... Now Plugging $A=0$ we get, $\sin 0*\cos 0*\tan 0 =0*1*0=0$ So, \Rightarrow (0+1+0)=\boxed{\color\red{1}}