Question

if sin (A-B) = 1/2 , cos(A+B) = 1/2 , 0° < A+B<90° , A>B , then find the value of A
plz....... i have to submit it morrow morning :'(

Answer 1

sinAcosB - sinBCosA = 1/2...........i

cosAcosB - sinAsinB = 1/2   .........ii

so  i divide ii
 sinAcosB - sinBCosA  = cosAcosB - sinAsinB
sinA(cosB + sinB) = CosA(sinB + CosB)
so tanA = 1
   A = 45°

Answer 2

Let sinAcosB - sinBCosA = 1/2 be equation (i) and
cosAcosB - sinAsinB = 1/2 be equation (ii)
Hence by dividing equation two by one we get;
sinAcosB - sinBCosA  = cosAcosB - sinAsinB
sinA(cosB + sinB) = CosA(sinB + CosB)
so tanA = 1
   A = 45°