If $$cotA=\frac{n}{(n+1)}$$ and $$cotB=\frac{1}{(2n+1)}$$, then what is the value of cot (A + B)?

Given : $$cotA=\frac{n}{(n+1)}$$ and $$cotB=\frac{1}{(2n+1)}$$

To find : $$cot(A+B)$$

= $$\frac{cotAcotB-1}{cotA+cotB}$$

= $$(\frac{n}{n+1}\times\frac{1}{2n+1}-1)\div(\frac{n}{n+1}+\frac{1}{2n+1})$$

= $$(\frac{n}{(n+1)(2n+1)}-1)\div(\frac{n(2n+1)+1(n+1)}{(n+1)(2n+1)})$$

= $$(\frac{n-(n+1)(2n+1)}{(n+1)(2n+1)})\div(\frac{2n^2+n+n+1}{(n+1)(2n+1)})$$

= $$\frac{n-(2n^2+3n+1)}{(n+1)(2n+1)}\times\frac{(n+1)(2n+1)}{2n^2+2n+1}$$

= $$\frac{-2n^2-2n-1}{2n^2+2n+1}=-1$$

=> Ans - (A)