Question 392592
using the exact values,find the value of: 

a.12 sin 30degrees - 6 tan 45 degress + sec 45 degrees / 52

b.sin 2A + tan 3A/2 - cos A + sec(A+15), when A= 30degrees


All you need to know is the value of the trigonometry functions at the special angles.  For this problem, you need to know the following:

sin 30 deg = 1/2
tan 45 deg = 1
sec 45 = sqrt(2)
cos 30 =sqrt(3)/2
sin 60 = sqrt(3)/2

a. 12 sin 30 deg - 6 tan 45 deg +sec 45 deg/52
    = 12*1/2-6*1+sqrt(2)/52=6-6+sqrt(2)/52
    =sqrt(2)/52

b.sin 2A + tan 3A/2 - cos A + sec(A+15)
   =sin (2*30)+tan (3*30/2)-cos 30+sec(30+15)
   = sin 60+tan 45-cos 30+sec 45
   = sqrt(3)/2+1-sqrt(3)/2+sqrt(2)
   = 1+sqrt(2)