frac{d}{dx} left( frac{sec x + tan x}{sec x - | vedclass

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Question

(A) Let $y = \frac{\sec x + 	an x}{\sec x - 	an x}$.Simplifying the expression: $y = \frac{\frac{1}{\cos x} + \frac{\sin x}{\cos x}}{\frac{1}{\cos x

Vedclass · Accepted Answer

$\frac{2\cos x}{(1 - \sin x)^2}$

Vedclass · Answer

(A) Let $y = \frac{\sec x + 	an x}{\sec x - 	an x}$.Simplifying the expression: $y = \frac{\frac{1}{\cos x} + \frac{\sin x}{\cos x}}{\frac{1}{\cos x} - \frac{\sin x}{\cos x}} = \frac{1 + \sin x}{1 - \sin x}$.Now,differentiate $y$ with respect to $x$ using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} ight) = \frac{v u' - u v'}{v^2}$:$u = 1 + \sin x \implies u' = \cos x$$v = 1 - \sin x \implies v' = -\cos x$$\frac{dy}{dx} = \frac{(1 - \sin x)(\cos x) - (1 + \sin x)(-\cos x)}{(1 - \sin x)^2}$$\frac{dy}{dx} = \frac{\cos x - \sin x \cos x + \cos x + \sin x \cos x}{(1 - \sin x)^2}$$\frac{dy}{dx} = \frac{2\cos x}{(1 - \sin x)^2}$.

$\frac{d}{dx} \left( \frac{\sec x + \tan x}{\sec x - \tan x} \right) = $

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