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Given that tan A - tan B = alpha and cot B - cot A = beta, and the expression (1 - cos 2(A-B)) / (1 + cos 2(A-B)) equals alpha^x * beta^y / (alpha + beta)^z, find the value of x + y - z.

1. 0
2. 1
3. 4
4. 2

Correct answer: 1

Solution

Since (1-cos2theta)/(1+cos2theta) = tan²(theta), the LHS equals tan²(A-B). One can show tan(A-B) = alpha*beta/(alpha+beta), so tan²(A-B) = alpha² * beta² / (alpha+beta)², giving x=2, y=2, z=2, and x+y-z = 2+2-2 = 2. Wait — re-checking: LHS = tan²(A-B), and alpha*beta = (tanA - tanB)(cotB - cotA) = (tanA-tanB)(1/tanB - 1/tanA) = (tanA-tanB)*(tanA-tanB)/(tanA*tanB) = alpha²/(tanA*tanB). Also alpha+beta = tanA - tanB + cotB - cotA = (tanA - cotA) - (tanB - cotB)... Let me redo: cot B - cot A = 1/tan B - 1/tan A = (tan A - tan B)/(tan A * tan B) = alpha/(tan A * tan B), so beta = alpha/(tan A * tan B), meaning tan A * tan B = alpha/beta. Now tan(A-B) = (tan A - tan B)/(1 + tan A * tan B) = alpha/(1 + alpha/beta) = alpha*beta/(alpha+beta). So tan²(A-B) = alpha² * beta² / (alpha+beta)², i.e. x=2, y=2, z=2, x+y-z = 2.

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