Find the value of $(1+\cot A-\csc A)(1+\tan A+\sec A)-3({\sin }^{2}A+{\cos }^{2}A)$.

Let's simplify the given expression:
$$\text{Expression}=(1+\cot A-\csc A)(1+\tan A+\sec A)-3({\sin }^{2}A+{\cos }^{2}A)$$

**Method 1: By substituting a value for $A$ **
Let $A=45^{\circ }$. We know that:

$$\tan 45^{\circ }=1,\cot 45^{\circ }=1$$

$$\sec 45^{\circ }=\sqrt{2},\csc 45^{\circ }=\sqrt{2}$$

$${\sin }^{2}45^{\circ }+{\cos }^{2}45^{\circ }=1$$

Substitute these values into the expression:

$$\text{Expression}=(1+1-\sqrt{2})(1+1+\sqrt{2})-3(1)$$

$$=(2-\sqrt{2})(2+\sqrt{2})-3$$

Using the identity $(a-b)(a+b)=a^2-b^2$:

$$=(2^2-(\sqrt{2}{)}^2)-3$$

$$=(4-2)-3$$

$$=2-3=-1$$

Method 2: Trigonometric Identity Reduction
We know that $(1+\cot A-\csc A)(1+\tan A+\sec A)=2$.
And from the standard Pythagorean identity: ${\sin }^{2}A+{\cos }^{2}A=1$.

$$\text{Expression}=2-3(1)=2-3=-1$$

Therefore, the required value is $-1$.