Question

PLEASE answer fast.......and correctly​

Answer 1

Answer:

We have,

sinA+sinB−−−−−−−−−−−+sinC=2sin(A+B2)cos(A−B2)+sinC....(⋆).

But, A+B+C=π∴A+B=π−C∴(A+B2)=π−C2.

∴(A+B2)=π2−C2.

∴sin(A+B2)=sin(π2−C2)=cos(C2)....(⋆1).

Also, sinC=2sin(C2)cos(C2).........(⋆2).

Utilising (

Answer 2

sinA+sinB-sinC

sinA+sinB-sinC= sinA + 2cos(B+C)/2 × sin(B-C)/2

= SinA + 2cos(π-A)/2 × sin(B-C)/2 [ Since A +B+C = π so B+C= π-A]

= SinA + 2cos(π/2-A/2) × sin(B-C)/2

= 2sinA/2×cosA/2 + 2sinA/2 × sin(B-C)/2

= 2sinA/2× [ cosA/2 + sin(B-C)/2]

= 2sinA/2× [ cos{ π/2 - (B+C)/2} + sin(B-C)/2]

[ Since A+B+C = π so A/2 = π/2 - (B+C)/2]

= 2sinA/2× [ sin (B+C)/2} + sin(B-C)/2]

= 2sin A/2 [ 2 sin 2B/2 cos 2C/2 ]

2B/2 cos 2C/2 ]= 2sin A/2 [ 2 sinB/2 cosC/2 ]

= 4 sin A/2 sinB/2 cosC/2 ]