Cos A sin(B-C) + cos B + cos C sin (A-B)=0

1.	Cos A sin(B-C) + cos B + cos C sin (A-B)=0
Answer» Step-by-step EXPLANATION: 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. Hope this HELPS!

Answer»

Step-by-step EXPLANATION:

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.

Hope this HELPS!

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Cos A sin(B-C) + cos B + cos C sin (A-B)=0

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