Question

if cosec a + cosec b + cosec c = 0
then prove that (sin a + sin b + sin c)² = sin²a + sin²b + sin²c.​

Answer 1

Step-by-step explanation:

given:

cosec a + cosec b + cosec c = 0

1/sin a + 1/sin b + 1/sin c = 0

sinb sin c + sina sin c + sin a sin b = 0-------(1)

now,

(sin a + sin b + sin c)(sin a + sin b + sin c)

= sin²a + 2sina sin b + 2sin a sinb + sin ²b + 2sin b sin c

= sin²a + sin²b + sin²c + 2(sinb sin c + sina sin c + sin a sin b)

= sin²a + sin²b + sin²c (using (1))

Answer 2

Answer:

(1/sinA)+(1/sinb)+(1/sinc,)=0

(sinb.sinc+sinA.sinc+sinA.sinb)/(sina.sinb.sinc)=0

sinb.sinc+sina.sinc+sina.sinb=0

(sina+sinb+sinc)^2=sin^2a+sin^2b+sin^2c +2(sina.sinb+sinb.sinc+sinc.sina)

=sin^2a+sin^2b+sin^2v+2×0