If TanA=k.TanB prove that (k+1)sin(A-B)=(k-1)sin(A+B).

k tan B = tan A

k sin B / cos B = sin A / cos A

k sin B cos A = sin A cos b

k (2 sin B cos A) = 2 sin A cos B

k (sin(A+B) - sin (A-B)) = sin (A-B) + sin(A+B)

k sin (A+B) - sin (A+B) = sin (A-B) + k sin (A-B)

(k-1) sin (A+B) = (k+1) sin (A-B)

Thanks for answer

Thanks for answer

Thanks

It is very useful for me thanku very much

Thanks a lot for the answer, much appreciated

If tanA =K tanB and A+B=C, prove that (K-1) sinC =(K+1) sin (A-B).

I just had a quick read of the response. The repsonse is excellent, however instead of multiplying both sides by 2 you could have just used the sinacosb and cosasinb identity immediately. This is a bit more complex, but I think it is easier to explain so those who are in doubt will be able to understand how you got to the solution.

I thin this is incorrect brudha

google your question, and you will find several solutions