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Question
If \[\tan \left( x + y \right) + \tan \left( x - y \right) = 1, \text{ find} \frac{dy}{dx}\] ?
Solution
\[\text{ We have, } \tan\left( x + y \right) + \tan\left( x - y \right) = 1\]
Differentiating with respect to x, we get,
\[\Rightarrow \frac{d}{dx}\tan\left( x + y \right) + \frac{d}{dx}\tan\left( x - y \right) = \frac{d}{dx}\left( 1 \right)\]
\[ \Rightarrow \sec^2 \left( x + y \right)\frac{d}{dx}\left( x + y \right) + \sec^2 \left( x - y \right)\frac{d}{dx}\left( x - y \right) = 0 \]
\[ \Rightarrow \sec^2 \left( x + y \right)\left[ 1 + \frac{dy}{dx} \right] + \sec^2 \left( x - y \right)\left[ 1 - \frac{dy}{dx} \right] = 0\]
\[ \Rightarrow \sec^2 \left( x + y \right)\frac{dy}{dx} - \sec^2 \left( x - y \right)\frac{dy}{dx} = - \left[ \sec^2 \left( x + y \right) + \sec^2 \left( x - y \right) \right]\]
\[ \Rightarrow \frac{dy}{dx}\left[ \sec^2 \left( x + y \right) - \sec^2 \left( x - y \right) \right] = - \left[ \sec^2 \left( x + y \right) + \sec^2 \left( x - y \right) \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sec^2 \left( x + y \right) + \sec^2 \left( x - y \right)}{\sec^2 \left( x - y \right) - \sec^2 \left( x + y \right)}\]
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