Question

please answer correctly.....and fats​

Answer 1

Answer:

or

you can solve by LHS and rhs formula

Answer 2

Answer:

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 ]

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

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