Simplify the following expression:
$$(1+{\tan }^{2}A)+\left(1+\frac{1}{{\tan }^{2}A}\right)$$

Given expression:
$$(1+{\tan }^{2}A)+\left(1+\frac{1}{{\tan }^{2}A}\right)$$

We know the standard trigonometric identities:

1. $1+{\tan }^{2}A={\sec }^{2}A$
2. $\frac{1}{{\tan }^{2}A}={\cot }^{2}A\Rightarrow 1+{\cot }^{2}A={\csc }^{2}A$

Substitute these identities back into the expression:

$${\sec }^{2}A+{\csc }^{2}A$$

Convert the terms into basic sine and cosine forms:

$$=\frac{1}{{\cos }^{2}A}+\frac{1}{{\sin }^{2}A}$$

Take the common denominator:

$$=\frac{{\sin }^{2}A+{\cos }^{2}A}{{\sin }^{2}A\cdot {\cos }^{2}A}$$

Since ${\sin }^{2}A+{\cos }^{2}A=1$:

$$=\frac{1}{{\sin }^{2}A\cdot {\cos }^{2}A}$$

Now substitute ${\cos }^{2}A=1-{\sin }^{2}A$ to match the denominator structure of the options:

$$=\frac{1}{{\sin }^{2}A(1-{\sin }^{2}A)}$$

$$=\frac{1}{{\sin }^{2}A-{\sin }^{4}A}$$