Respuesta :

If A is QIV, then        3π/2 ≤A≤2π; 
we have to find out in what quadrant is A/2
          (3π/2)/2≤A/2≤(2π)/2 ⇒ 3π/4≤A/2≤π
We can see that A/2 will be in QIII; therefore the sec (A/2) will be negative (-).

1) we have to calculate cos (A/2)
Cos (A/2)=⁺₋√[(1+cos A/2)/2]
We choose this formula: Cos (A/2)= -√[(1+cos A/2)/2],  because sec A/2 is in quadrant Q III, and the secant (sec A/2=1/cos A/2) in this quadrant is negative.

Cos (A/2)=-√[(1+cos A)/2]=-√[(1+(1/2)]/2=-√(3/4)=-(√3)/2.

2) we compute the sec (A/2)
Data:
cos (A/2)=-(√3)/2

sec (A/2)=1/cos (A/2)
sec (A/2)=1/(-(√3)/2)=-2/√3=-(2√3)/3

Answer: sec (A/2)=-(2√3)/3

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