Question

tan (B + C) + tan (C + A) + tan (A+B)/tan (π– A) + tan (2π- B) + tan (3π-C)=1
need help.......​

Answer 1

Step-by-step explanation:

since a+b+c = pi, a+b = pi-c, b+c = pi-a, c+a = pi-b

[tan(b+c)+tan(c+a)+tan(a+b)]/tan(pi-a)+tan(pi-b)+tan(pi-c)

[tan(pi-a)+tan(pi-c)+tan(pi-c)]/tan(pi-a)+tan(pi-b)+tan(pi-c) = 1