Question

If $\tan \left(x+y\right)+\tan \left(x-y\right)=1,\mathrm{find}\frac{dy}{dx}$

Answer 1

​$\mathrm{We}\mathrm{have},\tan \left(x+y\right)+\tan \left(x-y\right)=1$
Differentiating with respect to x, we get,
$$\begin{gathered} \Rightarrow \frac{d}{dx}\tan \left(x+y\right)+\frac{d}{dx}\tan \left(x-y\right)=\frac{d}{dx}\left(1\right) \\ \Rightarrow {\sec }^2\left(x+y\right)\frac{d}{dx}\left(x+y\right)+{\sec }^2\left(x-y\right)\frac{d}{dx}\left(x-y\right)=0 \\ \Rightarrow {\sec }^2\left(x+y\right)\left[1+\frac{dy}{dx}\right]+{\sec }^2\left(x-y\right)\left[1-\frac{dy}{dx}\right]=0 \\ \Rightarrow {\sec }^2\left(x+y\right)\frac{dy}{dx}-{\sec }^2\left(x-y\right)\frac{dy}{dx}=-\left[{\sec }^2\left(x+y\right)+{\sec }^2\left(x-y\right)\right] \\ \Rightarrow \frac{dy}{dx}\left[{\sec }^2\left(x+y\right)-{\sec }^2\left(x-y\right)\right]=-\left[{\sec }^2\left(x+y\right)+{\sec }^2\left(x-y\right)\right] \\ \Rightarrow \frac{dy}{dx}=\frac{{\sec }^2\left(x+y\right)+{\sec }^2\left(x-y\right)}{{\sec }^2\left(x-y\right)-{\sec }^2\left(x+y\right)} \end{gathered}$$