My answer isn't even in options. What am I doing wrong?
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There is an typo in the question. I google it and find "CBSE Sample Paper 2021-22". The correct question is: 
Now you can solve it.
$$\sin\left[\frac{\pi}3-\sin^{-1}\left(-\frac{1}2\right)\right]{=\sin\left[\frac{\pi}3+\sin^{-1}\left(\frac{1}2\right)\right]\\ =\sin\left[\frac{\pi}3+\frac{\pi}6\right]\\ =\sin \frac{\pi}{2}=1}$$
O M
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The set of all angles $\theta$ for which $\sin \theta = -\frac{1}{2}$ is $$\theta \in \left\{\ldots, -\frac{17\pi}{6}, -\frac{13\pi}{6}, -\frac{5\pi}{6}, -\frac{\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}, \ldots \right\} = \left\{-\frac{\pi}{6} + 2\pi k, -\frac{5\pi}{6} + 2\pi k\right\}_{k \in \mathbb Z}.$$ So the set $\frac{\pi}{2} - \theta$ is just $$\frac{\pi}{2} - \theta = \left\{\pm \frac{\pi}{3} + 2\pi k \right\}_{k \in \mathbb Z}.$$ Consequently $$\sin \left(\frac{\pi}{2} - \theta\right) = \pm \sin \frac{\pi}{3} = \pm \frac{\sqrt{3}}{2},$$ neither of which is among the included answer choices.
heropup
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But I had in mind that by definition, $|\sin^{-1} x|\leq \pi/2$ for all $x\in[-1,1]$. – Nothing special Nov 25 '23 at 06:03
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1@Nothingspecial It doesn't matter. Even with the broader criteria used in my answer, the answer is still not among the choices given. As the other answer has revealed, it is a typographical error and the correct expression is $\sin \left(\frac{\pi}{3} - \sin^{-1} \left(-\frac{1}{2}\right)\right)$. – heropup Nov 25 '23 at 06:32
$\frac{\sqrt{3}}{2}$to get $\frac{\sqrt{3}}{2}$. Feel free to ask any questions about MathJax. – Brian Tung Nov 25 '23 at 05:06$\sin^{-1}$gives you $\sin^{-1}$. ¶ Incidentally, I also get $\sqrt{3}/2$, so I'm not sure what's going on. – Brian Tung Nov 25 '23 at 05:10