(a) Prove that: (1) \( \cos (A+B)+\cos (A-B)=2 \cos A \cos B \) (3) \( \frac{\cos (x+y)-\sin (x-y)}{\cos 2 x}=\frac{\cos y+\sin y}{\cos x+\sin x} \)

**(1) Proof of \( \cos (A+B)+\cos (A-B)=2 \cos A \cos B \):** 1. Write the cosine sum and difference formulas: \[ \cos (A+B)=\cos A \cos B-\sin A \sin B \] \[ \cos (A-B)=\cos A \cos B+\sin A \sin B \] 2. Add the two identities: \[ \cos (A+B)+\cos (A-B) = (\cos A \cos B-\sin A \sin B)+(\cos A \cos B+\sin A \sin B) \] 3. The \(\sin A \sin B\) terms cancel out: \[ \cos (A+B)+\cos (A-B)=2\cos A \cos B \] --- **(3) Proof of \(\frac{\cos (x+y)-\sin (x-y)}{\cos 2 x}=\frac{\cos y+\sin y}{\cos x+\sin x}\):** 1. Express the numerator by applying the cosine and sine sum/difference formulas: \[ \cos (x+y)=\cos x\cos y-\sin x\sin y \] \[ \sin (x-y)=\sin x\cos y-\cos x\sin y \] 2. Substitute these into the numerator: \[ \cos (x+y)-\sin (x-y)= (\cos x\cos y-\sin x\sin y)-(\sin x\cos y-\cos x\sin y) \] 3. Distribute the negative sign and group like terms: \[ = \cos x\cos y-\sin x\sin y-\sin x\cos y+\cos x\sin y \] \[ = \cos x(\cos y+\sin y)-\sin x(\sin y+\cos y) \] \[ = (\cos x-\sin x)(\cos y+\sin y) \] 4. Express the denominator \( \cos 2x \) using the double-angle formula: \[ \cos 2x=\cos^2 x-\sin^2 x = (\cos x+\sin x)(\cos x-\sin x) \] 5. Substitute the expressions for numerator and denominator into the left-hand side: \[ \frac{\cos (x+y)-\sin (x-y)}{\cos 2 x} = \frac{(\cos x-\sin x)(\cos y+\sin y)}{(\cos x+\sin x)(\cos x-\sin x)} \] 6. Cancel the common factor \( \cos x-\sin x \) (assuming \( \cos x \neq \sin x \)): \[ = \frac{\cos y+\sin y}{\cos x+\sin x} \] Thus, the identity is proven.