Answer
see proof below.
Work Step by Step
Apply the sum formula:
$\tan(\alpha +\beta)=\dfrac{\tan \alpha + \tan \beta}{1 -\tan \alpha \ \tan \beta }$
In order to prove the given identity, we simplify the left hand side $\text{LHS}$ as follows:
$\text{LHS}=\cot (\alpha + \beta) \\=\dfrac{1}{\tan (\alpha + \beta)}
\\ =\dfrac{1}{\dfrac{\tan \alpha + \tan \beta}{1 -\tan \alpha \ \tan \beta }}$
Multiply the numerator and denominator by $\cot \alpha \cot \beta$ to obtain:
$\dfrac{\cot \alpha \cot \beta-\cot \alpha \cot \beta \times \tan \alpha \tan \beta} {\cot \alpha \cot \beta \ \tan \alpha-\cot \alpha \cot \beta \tan \beta}\\=\dfrac{\cot \alpha \ \cot \beta -1}{\cot \beta +\cot \alpha } \\=\text{ RHS}$
