Question

If A + B + C = π, and tan(A + B - C/4)tan (B + C - A/4) tan (A + C + B/4) = 1 prove that sinA + sinB + sinC + sinA sinB sinC = 0.

Answer 1

tan(A + B - C/4) = tan(π - 2C/4) = tan(π/4 - C/2) = 1 - tan C/2/1 + tan C/2

From Eqs. (1) and (2), we get

cosA cosB cosC = (1 − sinA) (1 − sinB) (1 − sinC)

= (1 + sinA) (1 + sinB) (1 + sinC)

Hence, 1 − Σ sinA + Σ sinA sinB − sinA sinB sinC

= 1 + Σ sinA + Σ sinA sinB + sinA sinB sinC

Therefore, Σ sinA + sinA sinB sinC = 0.