verify that cos(A-B)/cosAsinB=tanA+cotB is an identity
isnt Cos(A-B)= cosAcosB+sinASinB ?
To verify the given identity, we will simplify the left-hand side (LHS) and the right-hand side (RHS) and then show that they are equal.
Starting with the LHS, which is cos(A-B)/cos(A)sin(B):
Step 1: Apply the quotient-to-product identity for cosine:
cos(A-B)/cos(A)sin(B) = cos(A-B) / (cos(A)sin(B))
Step 2: Expand the numerator using the cosine difference formula:
cos(A-B) = cos(A)cos(B) + sin(A)sin(B)
Step 3: Replace cos(A-B) in the numerator with the expression from step 2:
(cos(A)cos(B) + sin(A)sin(B))/(cos(A)sin(B))
Step 4: Factor out common terms in the numerator:
cos(B)(cos(A)/sin(B)) + sin(A)
Step 5: Simplify the quotient cos(A)/sin(B) to cot(B):
cos(B)cot(B) + sin(A)
Moving on to the RHS, which is tan(A) + cot(B):
Since tan(A) is sin(A)/cos(A) and cot(B) is cos(B)/sin(B), we have:
tan(A) + cot(B) = sin(A)/cos(A) + cos(B)/sin(B)
Step 6: Find a common denominator:
[(sin(A)sin(B)) + (cos(A)cos(B))] / (cos(A)sin(B))
Step 7: Use the cosine sum identity:
[(sin(A)sin(B)) + (cos(A)cos(B))] / (cos(A)sin(B)) = cos(A-B)/cos(A)sin(B)
Comparing the LHS and RHS, we see that they are equal:
LHS = cos(B)cot(B) + sin(A) = tan(A) + cot(B) = RHS
Therefore, the given equation cos(A-B)/cos(A)sin(B) = tan(A) + cot(B) is an identity, as we have successfully verified it.