If $2\tan \beta +\cot \beta =\tan \alpha \text{ and }\cot \beta =n\tan (\alpha -\beta ),$ then n is equal to 

Given that,            $2\tan \beta +\cot \beta =\tan \alpha$             (i)

Now,             $\cot \beta =n\tan (\alpha -\beta )=n\cdot \frac{\tan \alpha -\tan \beta }{1+\tan \alpha \cdot \tan \beta }$

                                                   $\begin{array}{l}=n\cdot \frac{2\tan \beta +\cot \beta -\tan \beta }{1+(2\tan \beta +\cot \beta )\tan \beta }\text{ [trom Eq. (i)] }\\ =n\cdot \frac{\tan \beta +\cot \beta }{1+2{\tan }^2\beta +1}=n\cdot \frac{\tan \beta +\frac{1}{\tan \beta }}{2+2{\tan }^2\beta }\\ =n\cdot \frac{{\tan }^2\beta +1}{\tan \beta }\times \frac{1}{2\left(1+{\tan }^2\beta \right)}=\frac{n}{2}\cdot \frac{1}{\tan \beta }\end{array}$

$\begin{array}{r}\cot \beta =\frac{n}{2}\cdot \cot \beta \\ \therefore \frac{n}{2}=1\Rightarrow n=2\end{array}$