If $2 \tan B + \cot B = \tan A$, prove that $2 \tan (A-B) = \cot B$.

Question

If $2 \tan B + \cot B = \tan A$, prove that $2 \tan (A-B) = \cot B$. I am wondering how can i prove this question. If any one has answer of it then it would be a great pleasure.

Answer 1

It is: $$\begin{align}2 \tan (A-B) &= \frac{2\tan A-2\tan B}{1+\tan A\tan B}=\\ &=\frac{2(2\tan B+\cot B)-2\tan B}{1+(2\tan B+\cot B)\tan B}=\\ &=\frac{2(\tan B+\cot B)}{1+(2\tan B+\cot B)\frac1{\cot B}}=\\ &=\frac{2(\tan B+\cot B)\cot B}{2(\tan B+\cot B)}=\\ &=\cot B\end{align}$$