How do you simplify the expression $\dfrac{{{\tan }^{2}}x}{\sec x+1}$?

Hint: We know that we can write ${{\tan }^{2}}x$ as ${{\sec }^{2}}x-1$. Then we can apply the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ we can assume sec x as a and 1 as b then apply the formula to simplify the equation and we have to take care of the fact that denominator can not be equal to 0.

Complete step by step answer:
The given expression that we have simplify is $\dfrac{{{\tan }^{2}}x}{\sec x+1}$
We can see the numerator is tan x we can write square of tan x as ${{\sec }^{2}}x-1$
$\dfrac{{{\tan }^{2}}x}{\sec x+1}=\dfrac{{{\sec }^{2}}x-1}{\sec x+1}$
We know the algebraic formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
So now we can take sec x as a and 1 as b and apply the formula
$\Rightarrow \dfrac{{{\tan }^{2}}x}{\sec x+1}=\dfrac{\left( \sec x+1 \right)\left( \sec x+1 \right)}{\sec x+1}$
We can see that the denominator is equal to $\sec x+1$ and there is one $\sec x+1$ so we can cancel the terms
$\Rightarrow \dfrac{{{\tan }^{2}}x}{\sec x+1}=\sec x-1$ where sec x is not equal to -1 or x is not equal to $\left( 2n+1 \right)\pi $ where n is an integer

Note:
We can see that there is a mention at end of answer that sec x can not be -1 , this because we simplifies the equation by cancelling the term $\sec x+1$ , if the value of sec x is 0 then 0 can not be cancelled out, we can not cancel out 0 in numerator and denominator.
 The value of $\dfrac{{{\tan }^{2}}x}{\sec x+1}$ will not be equal to $\sec x-1$ when sec x is -1. The value $\sec x-1$is -2 when sec x is -1 but the value of $\dfrac{{{\tan }^{2}}x}{\sec x+1}$ is $\dfrac{0}{0}$ when sec x is -1. The value of $\dfrac{0}{0}$ is not defined.