find dy/dx

tan(x+y)+tan(x-y)=1

We have, \tan (x + y) + \tan (x - y) = 1 differentiating both sides with respect to x, we get (1 + \frac{d y}{d x}) s e c^{2} (x + y) + (1 - \frac{d y}{d x}) s e c^{2} (x - y) = 0 \Rightarrow \frac{d y}{d x} s e c^{2} (x + y) - \frac{d y}{d x} s e c^{2} (x - y) = - [s e c^{2} (x + y) + s e c^{2} (x - y)] \Rightarrow \frac{d y}{d x} = \frac{- [s e c^{2} (x + y) + s e c^{2} (x - y)]}{[s e c^{2} (x + y) - s e c^{2} (x - y)]} \Rightarrow \frac{d y}{d x} = \frac{[s e c^{2} (x + y) + s e c^{2} (x - y)]}{[s e c^{2} (x - y) - s e c^{2} (x + y)]}