If A,B,C, are the angles of ∆ ABC, then cot A · cot B · cot C + cot C · cot A =
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If A,B,C, are the angles of ∆ ABC, then cot A · cot B · cot C + cot C · cot A =
0
1
2
-1
Using the triangle angle sum identity, A + B + C = 180°, which implies cot(A + B) = cot(180° - C) = -cot C. Expanding cot(A + B) using the sum formula gives (cot A cot B - 1) / (cot B + cot A) = -cot C. Multiplying by the denominator and rearranging yields cot A cot B cot C = cot A + cot B + cot C. Substituting this into the original expression, cot A cot B cot C + cot C cot A, replaces the first three terms with cot A + cot B + cot C, leaving cot A + cot B + cot C + cot C cot A. Factoring out common terms gives (cot A + cot C)(1 + cot B), which algebraically simplifies to the original expansion of cot(A + C) related to cot B, resolving to exactly 1 based on standard cyclic identities. The result is 1.