Question

sin(A-B)/cosAcosB+sin(B-C)/cosBcosC+sin(C-A)/cosCcosA=0.Proof

Answer 1

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 2

hello users .....

we have to prove that 
Sin(A-B)/cosAcosB+sin(B-C)/cosBcosC+sin(C-A)/cosCcosA=0 

solution:-
formula's used:
sin (a -b) = sin a cos b - cos a sin b 

here 
Sin(A-B)/cosAcosB+sin(B-C)/cosBcosC+sin(C-A)/cosCcosA=

{(sin A cos B - cos A sin B ) / cos A cos B } + { (sin B cos C - cos C sin B) /

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 }

= sin A / cos A - sin B / cos B + sin B /cos B - sin C / cos C + sin C / cos C - sin A / cos A 

= tan A - tan B + tan B - tan C + tan C  - tan A 

= (tan A - tan A ) + ( tan B - tan B ) + ( tan C - tan C ) = 0 

hence;

Sin(A-B)/cosAcosB+sin(B-C)/cosBcosC+sin(C-A)/cosCcosA=0 

✪✵✪ hope it helps ✪✵✪