Question

show that
Sin(B-C)/
cosBcosC
+
sin(C-A)
CosCcosA
+
Sin (A-B)
CosAcosB
= 0​

Answer 1

Answer:

Given, sin(a-b)/cos a cos b + sin(b-c)/cos b cos c + sin(c-a)/cos c cos a

= sin a cos b - cos a sin b/cos a cos b + sin b cos c - cos b sin c/cos b cos c + sin c cos a - cos c sin a/cos c cos a

= sin a cos b/cos a cos b - cos a sin b/cos a cos b + sin b cos c/cos b cos c - cos b sinc/cos b cos c + sin c cos a/cos c cos a - cos c sin a/cos c cos a

= tan a - tan b + tan b - tan c + tan c - tan a

= 0