Answer
$\dfrac{\cot A-1}{1+\tan(-A)}=\cot A$
Work Step by Step
$\dfrac{\cot A-1}{1+\tan(-A)}$
Rewrite $\tan(-A)$ as $-\tan A$:
$\dfrac{\cot A-1}{1+\tan(-A)}=\dfrac{\cot A-1}{1-\tan A}=...$
Substitute $\cot A$ with $\dfrac{\cos A}{\sin A}$ and $\tan A$ with $\dfrac{\sin A}{\cos A}$:
$...=\dfrac{\dfrac{\cos A}{\sin A}-1}{1-\dfrac{\sin A}{\cos A}}=...$
Evaluate the substractions indicated in the numerator and in the denominator:
$...=\dfrac{\dfrac{\cos A-\sin A}{\sin A}}{\dfrac{\cos A-\sin A}{\cos A}}=...$
Evaluate the division and simplify:
$...=\dfrac{\cos A(\cos A-\sin A)}{\sin A(\cos A-\sin A)}=\dfrac{\cos A}{\sin A}=\cot A$
