Question

PLEASE answer it fats PLEASE don't answer irrelevant answers.......​

Answer 1

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

$\huge\red{Answer}$

______________________________

As you know that the sum of all angle of triangle is 180°

<A+<B+C=180°

B+C=180°-A

B+C/2=180/2-A/2 {if we dividing by 2 on both side }

sin²(B+C/2)=sin²(90°-A/2} {multiplying by sin²on both side,}

sin²(B+C)=cos²A/2 ........1)

now 1 substituting on given question

we get ..

sin²A/2+cos²A/2

=>1 ...like[ sin²A+cos²A=1]

hope it helps you

jai siya ram☺

______________________________

¯\_(ツ)_/¯