Answer
See the explanation below.
Work Step by Step
$\frac{{{\tan }^{2}}x-{{\cot }^{2}}x}{\tan x+\cot x}$
By using the formulae $(A+B)(A-B)={{A}^{2}}-{{B}^{2}}$ , with $A=\tan x$ and $B=\cot x$ for the first numeric expression, we get:
$\begin{align}
& \frac{{{\tan }^{2}}x-{{\cot }^{2}}x}{\tan x+\cot x}=\frac{\left( \tan x+\cot x \right)\left( \tan x-\cot x \right)}{\left( \tan x+\cot x \right)} \\
& =\tan x-\cot x
\end{align}$
Thus, the left side of the expression is equal to the right side, which is
$\frac{{{\tan }^{2}}x-{{\cot }^{2}}x}{\tan x+\cot x}=\tan x-\cot x$.
