If ABC is a triangle, then what is the value of the determinant |cosC sinB 0; tanA 0 sinB; 0 tan(B+C) cosC|?

In triangle ABC, A + B + C = π, so B + C = π − A, hence tan(B+C) = tan(π−A) = −tanA. Expanding the determinant along the first row: cosC·[0·cosC − sinB·tan(B+C)] − sinB·[tanA·cosC − sinB·0] + 0 = cosC·(−sinB·tan(B+C)) − sinB·tanA·cosC = −sinB·cosC·tan(B+C) − sinB·tanA·cosC = −sinB·cosC[tan(B+C) + tanA] = −sinB·cosC[−tanA + tanA] = 0.