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(D) Given expression: $E = a \cos (B-	heta) + b \cos (A+	heta)$ Using the expansion formula $\cos(x \pm y) = \cos x \cos y \mp \sin x \sin y$: $E =

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$c \cos 	heta$

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(D) Given expression: $E = a \cos (B-	heta) + b \cos (A+	heta)$ Using the expansion formula $\cos(x \pm y) = \cos x \cos y \mp \sin x \sin y$: $E = a(\cos B \cos 	heta + \sin B \sin 	heta) + b(\cos A \cos 	heta - \sin A \sin 	heta)$ $E = (a \cos B + b \cos A) \cos 	heta + (a \sin B - b \sin A) \sin 	heta$ By the projection formula in a triangle,$a \cos B + b \cos A = c$. By the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = 2R$,so $a = 2R \sin A$ and $b = 2R \sin B$. Substituting these into the second term: $a \sin B - b \sin A = (2R \sin A) \sin B - (2R \sin B) \sin A = 0$. Thus,$E = c \cos 	heta + 0 \cdot \sin 	heta = c \cos 	heta$.

With usual notations,in a triangle $ABC$,if $\theta$ is any real number,then $a \cos (B-\theta) + b \cos (A+\theta)$ is equal to:

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