Answer
The formulas are equivalent.
Work Step by Step
First, we manipulate $(\csc x+\cot x)$ by multiplying it with 1...
... and for 1, we choose $\displaystyle \frac{\csc x-\cot x}{\csc x-\cot x}$.
$\displaystyle \csc x+\cot x=(\csc x+\cot x)\cdot\frac{\csc x-\cot x}{\csc x-\cot x}=\frac{\csc^{2}x-\cot^{2}x}{\csc x-\cot x}$
... Pythagorean identity: $1+\cot^{2}x=\csc^{2}x,$
so, $\csc^{2}x-\cot^{2}x=1$
and
$\displaystyle \csc x+\cot x=\frac{1}{\csc x-\cot x}=(\csc x-\cot x)^{-1}$
Now, since the two LHS's are equal, we check if the RHS's are also equal.
The RHS of formula 1 =
$\ln|\csc x+\cot x|+C =\ln|\csc x-\cot x|^{-1}+C$
...Log Property: $\ln M^{n}=n\ln M...$
$=(-1)\ln|\csc x-\cot x|+C$
$... $= The RHS of formula 2
Therefore, the formulas are equivalent.