Prove that: \( \begin{array}{ll}\text { (1) } \cos (A+B)+\cos (A-B)=2 \cos A \cos B & \text { (2) } \cos \alpha \sin \beta=\frac{\sin (\alpha+\beta)-\sin (\alpha-\beta)}{2} \\ \text { (3) } \frac{\cos (x+y)-\sin (x-y)}{\cos 2 x}=\frac{\cos y+\sin y}{\cos x+\sin x} & \text { (4) } \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}\end{array} \)

We will prove each identity one by one. ────────────────────────────── (1) Prove that   cos(A + B) + cos(A − B) = 2 cos A cos B. Start with the sum-to-product formulas for cosine:   cos(A + B) = cos A cos B − sin A sin B,   cos(A − B) = cos A cos B + sin A sin B. Now, add the two expressions:   cos(A + B) + cos(A − B)     = [cos A cos B − sin A sin B] + [cos A cos B + sin A sin B]     = 2 cos A cos B. Thus, the first identity is proved. ────────────────────────────── (2) Prove that   cos α sin β = [sin(α + β) − sin(α − β)]⁄2. Recall the sine addition and subtraction formulas:   sin(α + β) = sin α cos β + cos α sin β,   sin(α − β) = sin α cos β − cos α sin β. Subtract the second formula from the first:   sin(α + β) − sin(α − β)     = [sin α cos β + cos α sin β] − [sin α cos β − cos α sin β]     = 2 cos α sin β. Dividing both sides by 2 gives:   cos α sin β = [sin(α + β) − sin(α − β)]⁄2. ────────────────────────────── (3) Prove that   [cos(x + y) − sin(x − y)]⁄cos 2x = [cos y + sin y]⁄[cos x + sin x]. Begin by expanding the numerator using angle addition formulas:   cos(x + y) = cos x cos y − sin x sin y   sin(x − y) = sin x cos y − cos x sin y. Therefore,   cos(x + y) − sin(x − y)     = [cos x cos y − sin x sin y] − [sin x cos y − cos x sin y]     = cos x cos y − sin x sin y − sin x cos y + cos x sin y     = cos x (cos y + sin y) − sin x (cos y + sin y)     = (cos x − sin x)(cos y + sin y). Now, look at the denominator cos 2x. Recall that:   cos 2x = cos²x − sin²x = (cos x − sin x)(cos x + sin x). Thus, we have:   [cos(x + y) − sin(x − y)]⁄cos 2x     = [(cos x − sin x)(cos y + sin y)]⁄[(cos x − sin x)(cos x + sin x)]     = (cos y + sin y)⁄(cos x + sin x),       (provided cos x − sin x ≠ 0). This proves the third identity. ────────────────────────────── (4) Prove that   tan(A + B) = [tan A + tan B]⁄[1 − tan A tan B]. Start with the sine and cosine addition formulas:   sin(A + B) = sin A cos B + cos A sin B,   cos(A + B) = cos A cos B − sin A sin B. Then,   tan(A + B) = sin(A + B)⁄cos(A + B)     = [sin A cos B + cos A sin B]⁄[cos A cos B − sin A sin B]. Now, divide numerator and denominator by cos A cos B (assuming cos A ≠ 0 and cos B ≠ 0):   tan(A + B) = [(sin A cos B)/(cos A cos B) + (cos A sin B)/(cos A cos B)]⁄[(cos A cos B)/(cos A cos B) − (sin A sin B)/(cos A cos B)]     = [tan A + tan B]⁄[1 − tan A tan B]. Thus, the fourth identity is verified. ────────────────────────────── Conclusion All four identities have been proved using standard trigonometric addition and subtraction formulas.