Giải njdjdndndndndndkdk

Ta có: `\hat(A) + \hat(B) + \hat(C) = 180^@`

`=> \hat(B) + \hat(C) = 180^@ - \hat(A)`

Do `cos A = - cos (180^@ - \hat(A))`

`=> cos A = - cos (B + C)`

Lại có:

`cot B + (cos C)/(sin B. cosA) = cot C + (cos B)/(sin C . cos A) (A \ne 90^@)`

`=> (cos B)/(sin B) + (cos C)/(sin B. cosA) = (cos C)/(sin C) + (cos B)/(sin C . cos A)`

`=> (cos B.cosA)/(sin B.cosA) + (cos C)/(sin B. cosA) = (cos C.cosA)/(sin C.cosA) + (cos B)/(sin C . cos A)`

`=> (cos B.cosA + cos C)/(sin B. cosA) = (cos C.cosA + cos B)/(sin C . cos A)`

`=> (cos B.cosA + cos C)/(sin B) = (cos C.cosA + cos B)/(sin C) (`nhân cả hai vế với `cos A \ne 0)`

`=> (- cos B.cos(B+C) + cos C)/(sin B) = (-cos C.cos(B+C) + cos B)/(sin C)`

`=> (- cos B.(cos B.cos C-sin B.sin C) + cos C)/(sin B) = (-cos C.(cos B.cos C-sin B.sin C) + cos B)/(sin C)`

`=> (- cos^2 B.cos C+cosB.sin B.sin C + cos C)/(sin B) = (-cos^2 C.cos B+sin B.sin C.cosC + cos B)/(sin C)`

`=> (cosC(1-cos^2 B)+cosB.sin B.sin C)/(sin B) = (cosB(1-cos^2 C)+sin B.sin C.cosC)/(sin C)`

`=> (cosC.sin^2 B+cosB.sin B.sin C)/(sin B) = (cosB.sin^2 C+sin B.sin C.cosC)/(sin C)`

`=> (sinB(cosC.sinB+cosB.sin C))/(sin B) = (sinC(cosB.sin C+sin B.cosC))/(sin C)`

`=> cosC.sinB+cosB.sin C = cosB.sin C+sin B.cosC` `(`luôn đúng`)`

Vậy `cot B + (cos C)/(sin B. cosA) = cot C + (cos B)/(sin C . cos A) (A \ne 90^@) (đpcm)`