. Prove that sin (A-B)/cos A cos B + sin (B-C)/cos B cos C+ sin (C-A)/cos C cos A=0


Best Answer

Explanation:

Taking Right hand side of the equation;

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 sin C/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

Same terms with opposite sign will cancel;

=0

= Left hand side of Equation

Final Answer:

Hence, sin (A-B)/cos A cos B+sin (B-C)/cos B cos C+sin (C-A)/cos C cos A=0

 

 

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