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(D) The given equation is $a \cos 	heta + b \sin 	heta = c$.Using the half-angle substitution $t = 	an(	heta/2)$,we have $\cos 	heta = \frac{1-t^

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$c = a$

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(D) The given equation is $a \cos 	heta + b \sin 	heta = c$.Using the half-angle substitution $t = 	an(	heta/2)$,we have $\cos 	heta = \frac{1-t^2}{1+t^2}$ and $\sin 	heta = \frac{2t}{1+t^2}$.Substituting these into the equation: $a(\frac{1-t^2}{1+t^2}) + b(\frac{2t}{1+t^2}) = c$.This simplifies to $a(1-t^2) + 2bt = c(1+t^2)$,which rearranges to $(c+a)t^2 - 2bt + (c-a) = 0$.Let $t_1 = 	an(\alpha/2)$ and $t_2 = 	an(\beta/2)$ be the roots of this quadratic equation.Then $t_1 + t_2 = \frac{2b}{c+a}$ and $t_1 t_2 = \frac{c-a}{c+a}$.If $\alpha + \beta$ is a root,then $	an(\frac{\alpha+\beta}{2})$ must satisfy the quadratic equation.Using the formula $	an(\frac{\alpha+\beta}{2}) = \frac{	an(\alpha/2) + 	an(\beta/2)}{1 - 	an(\alpha/2)	an(\beta/2)} = \frac{t_1+t_2}{1-t_1t_2} = \frac{2b/(c+a)}{1-(c-a)/(c+a)} = \frac{2b}{c+a-c+a} = \frac{2b}{2a} = \frac{b}{a}$.Substituting $t = b/a$ into $(c+a)t^2 - 2bt + (c-a) = 0$ gives $(c+a)(\frac{b^2}{a^2}) - 2b(\frac{b}{a}) + (c-a) = 0$.Multiplying by $a^2$: $(c+a)b^2 - 2b^2a + (c-a)a^2 = 0$.$cb^2 + ab^2 - 2ab^2 + ca^2 - a^3 = 0 \Rightarrow cb^2 - ab^2 + ca^2 - a^3 = 0$.$b^2(c-a) + a^2(c-a) = 0 \Rightarrow (c-a)(a^2+b^2) = 0$.Since $a, b$ are constants and the roots are distinct,$a^2+b^2 
eq 0$,so $c-a = 0$,which implies $c=a$.

Let $\alpha$ and $\beta$ be two distinct roots of $a \cos \theta + b \sin \theta = c$,where $a, b, c$ are three real constants and $\theta \in [0, 2\pi]$. Then,$\alpha + \beta$ is also a root of the same equation,if

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