# Answer to Question #12176 in Trigonometry for sathya krishna sai

Question #12176

prove that sin(A)-sin(B)/cos(A)+cos(B) + cos(A)-cos(B)/sin(A)+sin(B)=0

Expert's answer

[sin(A)-sin(B)]/[cos(A)+cos(B)] +

[cos(A)-cos(B)]/[sin(A)+sin(B)]=

=[sin^2(A)-sin^2(B)+

cos^2(A)-cos^2(B)]/[(sin(A)+sin(B))(cos(A)+cos(B))] =

=(1-1)/

[(sin(A)+sin(B))(cos(A)+cos(B))] =0

[cos(A)-cos(B)]/[sin(A)+sin(B)]=

=[sin^2(A)-sin^2(B)+

cos^2(A)-cos^2(B)]/[(sin(A)+sin(B))(cos(A)+cos(B))] =

=(1-1)/

[(sin(A)+sin(B))(cos(A)+cos(B))] =0

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