Question

tan(B-C) + tan(C-A) + tan(A-B) = tan(B-C)tan(C-A)tan(A-B)​

Answer 1

To show: tan(B-C)+tan(C-A)+tan(A-B)=tan(B-C)tan (C-A)tan(A-B)

solving tan(B-C)+tan(C-A)+tan(A-B)

tan(B-C)+tan(C-A)+tan(A-B)=tanB-tanC+tanC-tanA+tanA-tanB

tanB-tanC+tanC-tanA+tanA-tanB=0

Therefore, tan(B-C)+tan(C-A)+tan(A-B)=0

similarly, solving tan(B-C)tan (C-A)tan(A-B)

tan(B-C)tan(C-A)tan(A-B)=tanB-tanC+tanC-tanA+tanA-tanB

tanB-tanC+tanC-tanA+tanA-tanB=0

Therefore, tan(B-C) + tan(C-A) + tan(A-B) = tan(B-C)tan(C-A)tan(A-B)

Answer 2

Answer:

compare LHS and RHS

Step-by-step explanation:

Expand all the 3 terms of tan in LHS and RHS and compare