Let $f(x)=e^{\cos ^{-1}\{\sin (x+\pi / 3)\}}$. Then, $f(8 \pi / 9)=$

Question.
Let $f(x)=e^{\cos ^{-1}\{\sin (x+\pi / 3)\}}$. Then, $f(8 \pi / 9)=$
(a) $e^{5 \pi / 18}$
(b) $e^{13 \pi / 18}$
(c) $e^{-2 \pi / 18}$
(d) none of these

Solution:
(b) $e^{13 \pi / 18}$
Given: $f(x)=e^{\cos ^{-1}\{\sin (x+\pi / 3)\}}$
Then,
$f(8 \pi / 9)=e^{\cos ^{-1}\{\sin (8 \pi / 9+\pi / 3)\}}$
$=e^{\cos ^{-1}\{\sin (11 \pi / 9)\}}$
$=e^{\cos ^{-1}\left\{\cos \left(\pi / 2+{ }^{13 \pi} / 18\right)\right\}} \quad[\because \cos (\pi / 2+\theta)=\sin \theta]$
$=e^{\cos ^{-1}\{\cos (13 \pi / 18)\}}$
$=e^{13 \pi / 18}$

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