Question

Find the exact value of the trigonometric expression.

sec pi/12

The answer is √6-√2, but I'm not sure how to get it.

Answer 1

`\bf \textit{Half-Angle Identities} \\\\ sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}} \qquad cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}}\\\\ -------------------------------\\\\ \cfrac{1}{12}\cdot 2\implies \cfrac{1}{6}\qquad therefore\qquad \cfrac{\pi }{12}\cdot 2\implies \cfrac{\pi }{6}~thus~ \cfrac{\quad (\pi )/(6)\quad }{2}\implies \cfrac{\pi }{12}`


`\bf sec\left( (\pi )/(12) \right)\implies sec\left( \cfrac{(\pi )/(6)}{2} \right)\implies \cfrac{1}{cos\left( ((\pi )/(6))/(2) \right)}\impliedby \textit{now, let's do the bottom} \\\\\\ cos\left( \cfrac{(\pi )/(6)}{2} \right)=\pm\sqrt{\cfrac{1+cos\left( (\pi )/(6) \right)}{2}}\implies \pm\sqrt{\cfrac{1+(√(3))/(2)}{2}}\implies \pm\sqrt{\cfrac{(2+√(3))/(2)}{2}}`


`\bf \pm \sqrt{\cfrac{2+√(3)}{4}}\implies \pm\cfrac{\sqrt{2+√(3)}}{√(4)}\implies \pm \cfrac{\sqrt{2+√(3)}}{2} \\\\\\ therefore\qquad \cfrac{1}{cos\left( ((\pi )/(6))/(2) \right)}\implies \cfrac{2}{\sqrt{2+√(3)}}`


which simplifies thus far to


`\bf \cfrac{2}{\sqrt{2+√(3)}}\cdot \cfrac{\sqrt{2+√(3)}}{\sqrt{2+√(3)}}\implies \cfrac{2\sqrt{2+√(3)}}{2+√(3)}\implies \cfrac{2\sqrt{2+√(3)}}{2+√(3)}\cdot \stackrel{conjugate}{\cfrac{2-√(3)}{2-√(3)}} \\\\\\ \cfrac{2\sqrt{2+√(3)}(2-√(3))}{2^2-(√(3))^2}\implies \cfrac{2\sqrt{2+√(3)}(2-√(3))}{1}`


`\bf 2\sqrt{2+√(3)}(2-√(3))\impliedby \stackrel{\textit{keep in mind that}}{(2-√(3))=(\sqrt{2-√(3)})(\sqrt{2-√(3)})} \\\\\\ 2\sqrt{2+√(3)}\cdot \sqrt{2-√(3)}\cdot \sqrt{2-√(3)} \\\\\\ 2\sqrt{(2+√(3))(2-√(3))\cdot (2-√(3))} \\\\\\ 2\sqrt{[2^2-(√(3))^2]\cdot (2-√(3))}\implies 2\sqrt{1(2-√(3))} \\\\\\ 2\sqrt{2-√(3)}`