Question

show that sigma

(a+b)tan (a-b/2)=0
​

Answer 1

Answer:

a/sin A=b/sin B=c/sin C=k

(a-b)/(a+b)=(k sin A-k sin B)/(k sinA + k sinB).

or, (a-b)/(a+b)= (sin A - sin B)/(sin A+ sin B).

or, (a-b)/(a+b) = {2.cos (A+B)/2.sin(A-B)/2}/{2.sin(A+B)/2.cos(A-B)/2}.

or, (a-b)/(a+b) = cot(A+B)/2. tan(A-B)/2.

or, {(a-b)/(a+b)}= cot (90° -C/2).tan(A-B)/2.

or, {(a-b)/(a+b)} = tan C/2. tan (A-B)/2.

or, {(a-b)/(a+b)} = 1/cotC/2 . tan(A-B)/2.

or, {(a-b)/(a+b)}.cot C = tan(A-B)/2. Proved

Answer 2

Answer:

Hello here is ur answer

Step-by-step explanation: