Question

cosA+sinA/casA-sinA=?​

Answer 1

Step-by-step explanation:

cos A= sin(90°-A)

sinA+sinB=2[{sin(A+B) /2}*{cos(A-B)/2}]

sinA-sinB=2[{cos(A+B)/2}*{sin(A-B)/2}]

Now question

(sinA-cosA) /(cosA+sinA)

={sinA-sin(90°-A)}/{sin(90°-A)+sinA}

=2[{cos(A+90°-A)/2}*{sin(A-90°+A)/2}]/2[{sin(90°-A+A)/2}*{cos(90-A -A) /2}]

=[cos(90°/2)*sin(2A-90°)/2]/[sin(90°/2)*cos (90°-2A)/2]

=[cos45°*sin(2A-90°)/2]/[sin45°*cos(90°-2A)/2]

since sin45°=cos45°

=sin(A-45°)/cos(45°-A)

=sin(A-45°)/cos-(A-45°)

since cos(-A) =cosA

=tan(A-45°)