Question

Tan(x+y) +tan(x-y)=1 find dy/dx

Answer 1

Answer:

$\implies\sf (\frac{dy}{dx})= -[\frac{sec^2(x+y)+sec^2(x-y)}{(sec^2(x+y)-sec^2(x-y)}]$

Step-by-step explanation:

$\sf tan(x+y)+tan(x-y)=1$

differentiation wrt x

$\implies \sf sec^2(x+y)(1+\frac{dy}{dx})+sec^2(x-y) (1-\frac{dy}{dx})=0$

$\implies \sf sec^2(x+y)+sec^2(x+y)(\frac{dy}{dx})+sec^2(x-y) -sec^2(x-y)(\frac{dy}{dx})=0$

$\implies\sf sec^2(x+y)+sec^2(x-y)+sec^2(x+y)(\frac{dy}{dx})-sec^2(x-y)(\frac{dy}{dx})=0$

$\implies\sf sec^2(x+y)+sec^2(x-y)+(\frac{dy}{dx})(sec^2(x+y)-sec^2(x-y))=0$

$\implies\sf (\frac{dy}{dx})(sec^2(x+y)-sec^2(x-y))=-(sec^2(x+y)+sec^2(x-y))$

$\implies\sf (\frac{dy}{dx})= -[\frac{sec^2(x+y)+sec^2(x-y))}{(sec^2(x+y)-sec^2(x-y)}]$