Question

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

Answer 1

\[\text{ We have, } \tan\left( x + y \right) + \tan\left( x - y \right) = 1\]

Differentiating with respect to x, we get, 

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

\[ \Rightarrow \sec^2 \left( x + y \right)\frac{d}{dx}\left( x + y \right) + \sec^2 \left( x - y \right)\frac{d}{dx}\left( x - y \right) = 0 \]

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

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

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

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