Question

find DY\DX if tan(X + Y )+ tan(x - y )equal =1

Answer 1

that's your answer.....

Answer 2

Answer:

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

Step-by-step explanation:

Given : Expression $\tan(x+y)+\tan(x-y)=1$

To find : Solve for $\frac{dy}{dx}$?

Solution :

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

Differentiate w.r.t x both side,

$\frac{d}{dx}(\tan(x+y)+\tan(x-y))=\frac{dy}{dx}(1)$

$\sec^2(x+y)\cdot\frac{d}{dx}(x+y)+\sec^2(x-y)\cdot\frac{d}{dx}(x-y)=0$

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

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

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