2016/070) If $\tan(x+y) = a + b$ and $\tan(x-y) = a - b$ show that $a\tan\,x - b\tan\,y = a^2 - b^2$

$\tan(x+y) = \frac{\tan\,x+\tan\,y}{1-\tan\,x\tan\,y}$ or $(a+b)(1-\tan\,x\tan\,y) = \tan\,x+\tan\,y \cdots(1)$
Similarly $(a-b)(1+\tan\,x\tan\,y) = \tan\,x-\tan\,y\cdots(2)$
multilying (1) by (a-b) and (2) by (a+b) and adding we get $(a^2-b^2) =  2a\tan\,x-2b\tan\,y$
devide both sides by 2 to get the result.