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(C) Let $\alpha$ and $\beta$ be the integral roots of the quadratic equation $x^2 + (2 - 	an 	heta)x - (1 + 	an 	heta) = 0$.From the properties of

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$4$

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(C) Let $\alpha$ and $\beta$ be the integral roots of the quadratic equation $x^2 + (2 - 	an 	heta)x - (1 + 	an 	heta) = 0$.From the properties of roots,we have:$\alpha + \beta = 	an 	heta - 2$  $(1)$$\alpha \beta = -	an 	heta - 1$  $(2)$Adding $(1)$ and $(2)$,we get:$\alpha + \beta + \alpha \beta = -3$Adding $1$ to both sides to factorize:$\alpha \beta + \alpha + \beta + 1 = -3 + 1$$(\alpha + 1)(\beta + 1) = -2$Since $\alpha$ and $\beta$ are integers,$(\alpha + 1)$ and $(\beta + 1)$ must be integer factors of $-2$. The possible pairs for $(\alpha + 1, \beta + 1)$ are $(-1, 2), (2, -1), (1, -2), (-2, 1)$.Case $1$: $\alpha + 1 = -1$ and $\beta + 1 = 2 \Rightarrow \alpha = -2, \beta = 1$. Then $	an 	heta = \alpha + \beta + 2 = -2 + 1 + 2 = 1$.Case $2$: $\alpha + 1 = 1$ and $\beta + 1 = -2 \Rightarrow \alpha = 0, \beta = -3$. Then $	an 	heta = \alpha + \beta + 2 = 0 - 3 + 2 = -1$.For $	an 	heta = 1$ in $(0, 2\pi)$,$	heta = \frac{\pi}{4}, \frac{5\pi}{4}$.For $	an 	heta = -1$ in $(0, 2\pi)$,$	heta = \frac{3\pi}{4}, \frac{7\pi}{4}$.The sum of all possible values of $	heta$ is $\frac{\pi}{4} + \frac{5\pi}{4} + \frac{3\pi}{4} + \frac{7\pi}{4} = \frac{16\pi}{4} = 4\pi$.Thus,$k = 4$.

If the quadratic equation $x^2 + (2 - \tan \theta)x - (1 + \tan \theta) = 0$ has $2$ integral roots,then the sum of all possible values of $\theta$ in the interval $(0, 2\pi)$ is $k\pi$. Then $k$ equals:

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