If y=x+tan x, then cos ^2 x frac{d^2 y}{d x^2 | vedclass

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(D) Given, $y=x+	an x$. Differentiating with respect to $x$, we get: $\frac{dy}{dx} = 1 + \sec^2 x$. Differentiating again with respect to $x$, we ge

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2y

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(D) Given, $y=x+	an x$. Differentiating with respect to $x$, we get: $\frac{dy}{dx} = 1 + \sec^2 x$. Differentiating again with respect to $x$, we get: $\frac{d^2y}{dx^2} = 0 + 2 \sec x \cdot (\sec x 	an x) = 2 \sec^2 x 	an x$. Now, substitute this into the expression $\cos^2 x \frac{d^2y}{dx^2} + 2x$: $\cos^2 x (2 \sec^2 x 	an x) + 2x$. Since $\cos^2 x \cdot \sec^2 x = 1$, the expression simplifies to: $2 	an x + 2x = 2(x + 	an x)$. Since $y = x + 	an x$, the expression is equal to $2y$.

If $y=x+\tan x$, then $\cos ^2 x \frac{d^2 y}{d x^2}+2 x$ is equal to

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