Question

In a ∆ABC ,b cos (C+teta)+c cos(B-teta)​

Answer 1

Answer:

bcos(A−θ)+acos(B+θ)

=b(cosA.cosθ+sinA.sinθ)+a(cosB.cosθ−sinB.sinθ)

=cosθ(bcosA+acosB)+sinθ(bsinA−asinB)

=cosθ(c)+sinθ(b

2R

a

−a

2R

b

) [Since acosB+bcosA=c and using sine law of triangle]

=ccosθ.