Question

prove that a-b/c=(sin(A_B)/2)/cos c/2​



pls send the answer fast....its urgent....

Answer 1

Step-by-step explanation:

From sine law, we have, asinA=bsinB=csinC=k

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

∴L.H.S.=a−bc=k(sinA−sinB)ksinC

=sinA−sinBsinC

=2sin(A−B2)cos(A+B2)2sin(C2)cos(C2)

Here, A+B+C=180.So,A+B=180−C

So, our expression becomes,

=sin(A−B2)cos(180−C2)sin(C2)cos(C2)

=sin(A−B2)cos(90−C2)sin(C2)cos(C2)

=sin(A−B2)sin(C2)sin(C2)cos(C2)

=sin(A−B2)cos(C2)=R.H.S.

Answer 2

Answer:

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.