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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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} = $

Similar Questions

If $x^y \cdot y^x = 16$,then $\frac{dy}{dx}$ at $(2, 2)$ is

The equation of the normal to the curve $y=(1+x)^{2y}+\cos^{2}(\sin^{-1} x)$ at $x=0$ is

If $y=1+xe^y$,then $\frac{dy}{dx}=$

If $x \sqrt{1+y}+y \sqrt{1+x}=0$ for $-1 < x < 1$,prove that $\frac{dy}{dx} = -\frac{1}{(1+x)^2}$.

If $x^y=y^x$,then $x(x-y \log x) \frac{d y}{d x}$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module