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(C) Given the equation $a \cos x + b \sin x = c.$Using the half-angle formulas $\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$ and $\sin x = \frac{

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$\frac{b}{a}$

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(C) Given the equation $a \cos x + b \sin x = c.$Using the half-angle formulas $\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$ and $\sin x = \frac{2 	an(x/2)}{1 + 	an^2(x/2)},$ we get:$a \left( \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)} ight) + b \left( \frac{2 	an(x/2)}{1 + 	an^2(x/2)} ight) = c$$a(1 - 	an^2(x/2)) + 2b 	an(x/2) = c(1 + 	an^2(x/2))$$(a + c) 	an^2(x/2) - 2b 	an(x/2) + (c - a) = 0$Let $t = 	an(x/2).$ The roots of this quadratic equation are $t_1 = 	an(\alpha/2)$ and $t_2 = 	an(\beta/2).$From the properties of quadratic equations,the sum of roots is $t_1 + t_2 = \frac{2b}{a + c}$ and the product of roots is $t_1 t_2 = \frac{c - a}{a + c}.$We need to find $	an \left( \frac{\alpha + \beta}{2} ight) = 	an \left( \frac{\alpha}{2} + \frac{\beta}{2} ight).$Using the identity $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B},$$	an \left( \frac{\alpha + \beta}{2} ight) = \frac{t_1 + t_2}{1 - t_1 t_2} = \frac{\frac{2b}{a + c}}{1 - \frac{c - a}{a + c}} = \frac{2b}{a + c - c + a} = \frac{2b}{2a} = \frac{b}{a}.$

If $\alpha$ and $\beta$ are different values of $x$ satisfying $a \cos x + b \sin x = c,$ then $\tan \left( \frac{\alpha + \beta}{2} \right) = $

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