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

Use the standard trigonometric identities:
1. $1+{\tan }^{2}A={\sec }^{2}A$
2. $1+\frac{1}{{\tan }^{2}A}=1+{\cot }^{2}A={\csc }^{2}A$

Substitute these back into the expression:

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

Convert them into terms of sine and cosine:

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

Since ${\sin }^{2}A+{\cos }^{2}A=1$, the expression becomes:

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

Substitute ${\cos }^{2}A=1-{\sin }^{2}A$ into the denominator:

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