If 9cosA + 40 sinA =41 .cosA and cosecA = ?

1.	If 9cosA + 40 sinA =41 .cosA and cosecA = ?
Answer» i) Dividing both sides by 41, (9/41)cos(A) + (40/41)sin(A) = 1 ii) This is of the form, cos(A)cos(B) + sin(A)sin(B) = 1, where cos(B) = 9/41 and sin(B) = 40/41 iii) So, cos(A - B) = 1 ==> A - B = 2nπ Hence A = 2nπ + B iv) So, cos(A) = cos(2nπ + B) = cos(B) = 9/41 csc(A) = csc(2nπ + B) = csc(B) = 41/40

Answer»

i) Dividing both sides by 41, (9/41)*cos(A) + (40/41)*sin(A) = 1

ii) This is of the form, cos(A)*cos(B) + sin(A)*sin(B) = 1, where cos(B) = 9/41 and sin(B) = 40/41

iii) So, cos(A - B) = 1 ==> A - B = 2nπ Hence A = 2nπ + B

iv) So, cos(A) = cos(2nπ + B) = cos(B) = 9/41 csc(A) = csc(2nπ + B) = csc(B) = 41/40

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