Prove the following trigonometric function: $1+\dfrac{{{\tan }^{2}}A}{1+\sec A}=\sec A$.

Hint: We will be using the concept of trigonometric function and using trigonometric identities ${{\sec }^{2}}A-{{\tan }^{2}}A=1$ and basic algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ for solving the question.

Complete step-by-step answer:
Now, we have to prove that,
$1+\dfrac{{{\tan }^{2}}A}{1+\sec A}=\sec A$
Now, we will first take the left-hand side of the equation and prove it to be equal to the right hand side. In left hand side we have,
$1+\dfrac{{{\tan }^{2}}A}{1+\sec A}$
Now, we know the trigonometric identity that,
$\begin{align}
  & {{\sec }^{2}}A-{{\tan }^{2}}A=1 \\
 & \Rightarrow {{\sec }^{2}}A-1={{\tan }^{2}}A.............\left( 1 \right) \\
\end{align}$
Now, we will substitute the value of ${{\tan }^{2}}A$ from (1) in $1+\dfrac{{{\tan }^{2}}A}{1+\sec A}$ .
So, we have,
$1+\dfrac{{{\tan }^{2}}A}{1+\sec A}=1+\dfrac{{{\sec }^{2}}A-1}{1+\sec A}$
Now, we know the algebraic identity that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$. Therefore, we have,
$=1+\dfrac{\left( \sec A-1 \right)\left( \sec A+1 \right)}{\left( 1+\sec A \right)}$
Now, $1+\sec A$ will get canceled in both numerator and denominator. So, we have,
$\begin{align}
  & =1+\sec A-1 \\
 & 1+\dfrac{{{\tan }^{2}}A}{1+\sec A}=\sec A \\
\end{align}$
Since, L.H.S. = R.H.S. Therefore, we have proved that,
$1+\dfrac{{{\tan }^{2}}A}{1+\sec A}=\sec A$.

Note: To solve these type of question it is important to remember trigonometric identities and formulas like ${{\sec }^{2}}A-{{\tan }^{2}}A=1$, ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ and basic algebraic identities like ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ for solving these types of questions.