Question

PLEASE answer it fast and don't answer irrelevantly please ​

Answer 1

Step-by-step explanation:

given A+B+C=180

LHS      SIN²A/2+SIN²B/2+SIN²C/2

            1/2(2SIN²A/2+2SIN²B/2+2SIN²C/2)

            1/2(1-COSA+1-COSB+1-COSC)

            1/2(3-(COSA+COSB+COSC)

            1/2(3- (2COS (A+B/2)COS(A-B/2)+COSC)

            1/2(3- (2COS(90-C/2)COS(A+B/2)+C0SC)

            1/2(3- (2SINC/2COS(A-B/2)+1-2SIN²C/2)

            1/2(3-1-2SINC/2(COS(A+B/2)-SINC/2)

            1/2(2-2SINC/2(COS(A-B/2)-SIN (90-A+B/2)  (SIN90-Ф)= COSФ)

            1/2(2-2SINC/2(COS(A-B/2)-COS(A+B/2)

    /2(2-2SINC/2(2SINA/2SINB/2)            1/2(2-4SINA/2SINB/2SINC/2)            1-2SINA/2SINB/2SINC/2            RHS

Answer 2

HOLA!!

HERE IS UR ANSWER DEAR:--

SEE THE ATTACHMENT FOR DETAILED SOLUTION TO THE QUESTION DEAR ✔️⚡⚡

MARK MY ANSWERS BRAINLIAST AND FOLLOW ME ✌