Answer
$\dfrac{2}{\sqrt{{2+\sqrt 2}}}$
Work Step by Step
The inverse Identity for secant can be expressed as:
$\sec a =\dfrac {1}{\cos a}$ or, $\sec a =(\cos a)^{-1}$
Therefore,
$\sec ({\dfrac{15 \pi }{8}})=( \sqrt{\dfrac{1 +\cos ({\dfrac{15 \pi }{8}})}{2}} )^{-1} \\=[ \sqrt{\dfrac{1 +\dfrac{\sqrt 2}{2}}{2}}]^{-1} \\=[\dfrac{\sqrt {2+\sqrt 2}}{2}]^{-1}\\ = \dfrac{2}{\sqrt{{2+\sqrt 2}}}$
