Question 32484
TAN(A+B)=SIN(A+B)/COS(A+B)
={SIN(A)COS(B)+COS(A)SIN(B)}/{COS(A)COS(B)-SIN(A)SIN(B)}...DIVIDING THROUGHOUT BY COS(A)COS(B) WE GET THIS
=[{SIN(A)COS(B)/COS(A)COS(B)}+{COS(A)SIN(B)/COS(A)COS(B)}]/[{COS(A)COS(B)/COS(A)COS(B)}-{SIN(A)SIN(B)/COS(A)COS(B)]
={TAN(A)+TAN(B)}/{1-TAN(A)TAN(B)}