Let $f (x) = e^{\cos^{- 1} \sin (x + π / 3)}$ then

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$f (\frac{8 π}{9}) = e^{13 π / 18}$

$f (\frac{8 π}{9}) = e^{5 π / 18}$

$f (- \frac{7 π}{4}) = e^{π / 12}$

$f (- \frac{7 π}{4}) = e^{11 π / 12}$

answer is B, C.

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Detailed Solution

$\begin{array}{l} f (x) = e^{\cos^{- 1} \sin (x + π / 3)} \\ f (\frac{8 π}{9}) = e^{\cos^{1} (\sin (\frac{11 π}{9}))} \\ \Rightarrow f (\frac{8 π}{9}) = e^{\cos^{1} (\sin (\frac{π}{2} + \frac{13 π}{18}))} \\ \Rightarrow f (\frac{8 π}{9}) = e^{\cos^{1} \cos \frac{13 π}{18}} = e^{\frac{13 π}{18}} \\ \Rightarrow f (\frac{- 7 π}{4}) = e^{\cos^{1} \sin (\frac{- 7 π}{4} + \frac{π}{3})} = e^{\cos^{1} \sin (\frac{- 17 π}{12})} \\ \Rightarrow f (\frac{- 7 π}{4}) = e^{\cos^{1} \sin (\frac{5 π}{12})} \\ \Rightarrow f (\frac{- 7 π}{4}) = e^{\cos^{1} \sin (\frac{π}{2} - \frac{π}{12})} \\ \Rightarrow f (\frac{- 7 π}{4}) = e^{\cos^{1} \cos \frac{π}{12}} = e^{\frac{π}{12}} \end{array}$