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(C) Given the curve $y = \cos(x+y)$.Taking the derivative with respect to $x$:$\frac{dy}{dx} = -\sin(x+y) \cdot (1 + \frac{dy}{dx})$$\frac{dy}{dx} (1

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$\frac{\pi}{2}$

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(C) Given the curve $y = \cos(x+y)$.Taking the derivative with respect to $x$:$\frac{dy}{dx} = -\sin(x+y) \cdot (1 + \frac{dy}{dx})$$\frac{dy}{dx} (1 + \sin(x+y)) = -\sin(x+y)$$\frac{dy}{dx} = \frac{-\sin(x+y)}{1 + \sin(x+y)}$The equation of the tangent is $x + 2y = k$, which can be written as $y = -\frac{1}{2}x + \frac{k}{2}$.The slope of the tangent is $m = -\frac{1}{2}$.Equating the slopes: $\frac{-\sin(x+y)}{1 + \sin(x+y)} = -\frac{1}{2}$$2\sin(x+y) = 1 + \sin(x+y)$$\sin(x+y) = 1$Since $\sin(x+y) = 1$, we have $y = \cos(x+y) = \cos(\frac{\pi}{2} + 2n\pi) = 0$.Substituting $y = 0$ into $\sin(x+y) = 1$, we get $\sin(x) = 1$, so $x = \frac{\pi}{2}$.Now, substitute $x = \frac{\pi}{2}$ and $y = 0$ into the tangent equation $x + 2y = k$:$\frac{\pi}{2} + 2(0) = k$$k = \frac{\pi}{2}$.

If the equation of a tangent drawn to the curve $y=\cos (x+y), -1 \leq x \leq 1+\pi$ is $x+2y=k$, then $k=$

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