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\frac { \operatorname { sin } ( B - C ) } { \operatorname { cos } B \operatorname { cos } C } + \frac { \operatorname { sin } ( C - A ) } { \operatorname { cos } C \operatorname { cos } A } + \frac { \operatorname { sin } ( A - B ) } { \operatorname { cos } A \operatorname { cos } B } = 0 |
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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. |
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