Solution :
We know that \(sin^{-1}(sinx)\) = x, if \(-\pi\over 2\) \(\le\) x \(\le\) \(\pi\over 2\)
Here, x = 10 radians which does not lie between -\(\pi\over 2\) and \(\pi\over 2\)
But, \(3\pi\) – x i.e. \(3\pi\) – 10 lie between -\(\pi\over 2\) and \(\pi\over 2\)
Also, sin(\(3\pi\) – 10) = sin 10
\(\therefore\) \(sin^{-1}(sin10)\) = \(sin^{-1}(sin(3\pi – 10)\) = (\(3\pi\) – 10)
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