Question 535813
Find the exact value of sec (71[pi]/12) 
the answer is (root 6 - root 2) but i don't know how to get to the answer
**
Finding the reference angle:
Notice that the given angle is &#960;/12 short of 72&#960;/12 or 6&#960;, the origin; so the reference angle is &#960;/12 in quadrant IV where sec and cos>0 and sin<0
sec(&#960;/12)=1/cos(&#960;/12)
cos(&#960;/12)=cos(&#960;/3-&#960;/4)=cos(4&#960;/12-3&#960;/12)
Using cos addition formula: Cos(s-t)=cos s cos t+sin s sin t
cos(&#960;/3+&#960;/4)=cos(&#960;/3)cos(&#960;/4)+sin(&#960;/3)sin(&#960;/4)
cos(&#960;/3+&#960;/4)=1/2*&#8730;2/2+&#8730;3/2*&#8730;2/2)=&#8730;2/4+&#8730;6/4=(&#8730;2+&#8730;6)/4)
cos(&#960;/3+&#960;/4)=(&#8730;2+&#8730;6)/4)
Take reciprocal
sec(&#960;/3+&#960;/4)=4/(&#8730;2+&#8730;6)
rationalize denominator of 4/(&#8730;2+&#8730;6)
multiply numerator and denominator by &#8730;2-&#8730;6
4(&#8730;2-&#8730;6)/(&#8730;2+&#8730;6)(&#8730;2-&#8730;6) (makes denominator difference of squares)
4(&#8730;2-&#8730;6)/(2-6)
4(&#8730;2-&#8730;6)/-4
&#8730;6-&#8730;2
Ans:
sec (71[pi]/12)=&#8730;6-&#8730;2