Question

For a triangle ABC, prove that a+b/c=Cos(A-B/2)/SinC/2

Answer 1

The sum of all angles in a triangle is 180°
Therefore,
A + B + C = 180°
=> A + B = 180° - C
=> (A + B)/2 = (180° - C)/2 = 90° - C/2
In the above question,
LHS = cos((A + B)/2)
substituting (A + B)/2 with 90°- C/2
= cos(90° - C/2)
cos(90° - θ) = sinθ
Therefore,
= sin(C/2) which is also = RHS
LHS = RHS
Hence proved.

Answer 2

Answer:

Step-by-step explanation:

Using sine formula ,

a/sinA =b/sinB =c/sinC =k(say)

,a =ksinA, b =ksinB, c =ksinC

LHS

(a+b)/c = (ksinA +ksinB)/ksinC

=k(2.sin(A+B)/2).cos(A-B)/2)

/(2.sinC/2.cosC/2)

= (sin(π/2-C/2).cos(A-B/2))

/(sinC/2.cosC/2)

=(cosC/2.cos(A-B/2))

/(sinC/2.cosC/2

=cos((A-B)/2)/sin(C/2)=RHS