Question:

<p><strong>Step 1: Simplify the given expression</strong></p><p>Given: $y = \frac{\sec x + \tan x}{\sec x - \tan x}$</p><p>Multiply both numerator and denominator by $(\sec x + \tan x)$:</p><p>$$y = \frac{(\sec x + \tan x)^2}{(\sec x - \tan x)(\sec x + \tan x)} = \frac{(\sec x + \tan x)^2}{\sec^2 x - \tan^2 x}$$</p><p><strong>Step 2: Use the trigonometric identity</strong></p><p>Recall that $\sec^2 x - \tan^2 x = 1$</p><p>Therefore:</p><p>$$y = (\sec x + \tan x)^2$$</p><p><strong>Step 3: Expand the expression</strong></p><p>$$y = \sec^2 x + 2\sec x \tan x + \tan^2 x$$</p><p><strong>Step 4: Differentiate with respect to x</strong></p><p>$$\frac{dy}{dx} = 2\sec x(\sec x \tan x) + 2\sec x \tan x + 2\sec x \tan x \sec x + 2\tan x \sec^2 x$$</p><p>Using derivatives: $\frac{d}{dx}(\sec x) = \sec x \tan x$ and $\frac{d}{dx}(\tan x) = \sec^2 x$</p><p>$$\frac{dy}{dx} = 2\sec x \tan x (\sec x + \tan x) + 2\sec^2 x(\sec x + \tan x)$$</p><p>$$\frac{dy}{dx} = 2(\sec x + \tan x)(\sec x \tan x + \sec^2 x)$$</p><p>$$\frac{dy}{dx} = 2\sec x(\sec x + \tan x)(\tan x + \sec x)$$</p><p>$$\frac{dy}{dx} = 2\sec x(\sec x + \tan x)^2$$</p><p>Alternatively, directly differentiating $y = (\sec x + \tan x)^2$:</p><p>$$\frac{dy}{dx} = 2(\sec x + \tan x) \cdot \frac{d}{dx}(\sec x + \tan x)$$</p><p>$$= 2(\sec x + \tan x)(\sec x \tan x + \sec^2 x)$$</p><p>$$= 2(\sec x + \tan x) \sec x(\tan x + \sec x)$$</p><p>$$= 2\sec x(\sec x + \tan x)$$</p><p><strong>∴ Answer: A</strong></p>