Question

if 9cos A + 40sin A = 41 then find the values of cosA and cosec A

Answer 1

 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 2

Given 9cosA+40sinA=41
we know that 9 cosA- 40sinA=1/41
by solving two equations
we get 18 cosA=41+1/41
cosA =1681/41
By subtracting two equations we get
80sinA=41-1/41
sin A=1680/41*80
sinA= 21/41
cosecA=41/21