Question

If  $$\tan (x+y)+\tan (x-y)=1,\text{ find}\frac{dy}{dx}$$ ?

Answer 1

$$\text{ We have, }\tan (x+y)+\tan (x-y)=1$$

Differentiating with respect to x, we get, 

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

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

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

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

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

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