If tan (x + y) + tan (x - y) = 1, then frac{d | vedclass

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(B) Given the equation: $	an (x + y) + 	an (x - y) = 1$Differentiating both sides with respect to $x$ using the chain rule:$\sec^2(x + y) \cdot \fra

Vedclass · Accepted Answer

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

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(B) Given the equation: $	an (x + y) + 	an (x - y) = 1$Differentiating both sides with respect to $x$ using the chain rule:$\sec^2(x + y) \cdot \frac{d}{dx}(x + y) + \sec^2(x - y) \cdot \frac{d}{dx}(x - y) = 0$$\sec^2(x + y) \left(1 + \frac{dy}{dx}ight) + \sec^2(x - y) \left(1 - \frac{dy}{dx}ight) = 0$Expanding the terms:$\sec^2(x + y) + \sec^2(x + y) \frac{dy}{dx} + \sec^2(x - y) - \sec^2(x - y) \frac{dy}{dx} = 0$Grouping the $\frac{dy}{dx}$ terms:$\frac{dy}{dx} \left( \sec^2(x + y) - \sec^2(x - y) ight) = - \left( \sec^2(x + y) + \sec^2(x - y) ight)$Solving for $\frac{dy}{dx}$:$\frac{dy}{dx} = - \frac{\sec^2(x + y) + \sec^2(x - y)}{\sec^2(x + y) - \sec^2(x - y)}$$\frac{dy}{dx} = \frac{\sec^2(x + y) + \sec^2(x - y)}{\sec^2(x - y) - \sec^2(x + y)}$Thus,the correct option is $(b)$.

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

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