Question

A,B,C are angle of a triangle then
find sin (A+B/2) ii) cosec (B+C/2). pls answer to this question ..

i will mark you as brainliest​

Answer 1

Answer:

1) sin (A+B/2) = cos C/2

2)cosec (B+C/2) =sec (A/2)

sum of angles of triangle = 180degrees

A+B+C = 180

1) sin (A+B/2) = cos C/2

A+B+C = 180

A+B =180-C

sin(180-C/2)

sin (90-C/2) .[sin (90-ô) = cosô]

cos C/2

2)cosec (B+C/2) =cot (A/2)

A+B+C = 180

B+C = 180-A

cosec ( 180-A/2)

cosec ( 90-A/2) [cosec(90-ô) = secô]

sec (A/2)

Answer 2

Answer:

On solving, we get

Sin(A+B/2) = Cos(C/2)

&

Cosec(B+C/2) = Sec(A/2)