`Delta ABC` में सिद्ध कीजिए कि `a cos.(B-C)/(2)=(b+c)sin.(A)/(2)`

Explanation

हम जानते है कि `(sinA)/(A)=(sinB)/(b)=(sinC)/(c)=(1)/(k)`
`rArr a=k sin A, b=k sin B, c= sin C`
अब
`(b+c)/(a)=(ksinB+ksinC)/(ksinA)`
`=(sinB+sinC)/(sinA)=(2sin((B+C)/(2))cos((B-C)/(2)))/(2sin.(A)/(2)cos.(A)/(2))`
`=(sin(90^@-(A)/(2))cos((B-C)/(2)))/(sin.(A)/(2)cos.(A)/(2))=(cos.(A)/(2)cos((B-C)/(2)))/(sin.(A)/(2)cos.(A)/(2))=(cos.(B-C)/(2))/(sin.(A)/(2))`
इसलिए `a cos.(B-C)/(2)=(b+c)sin.(A)/(2)` .