sumlimits_{lambda = 1}^{10} {{{sin }^{ - 1}}l | vedclass

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(C) Let $S = \sum\limits_{\lambda = 1}^{10} {{{\sin }^{ - 1}}\left( {\sin \left( {\lambda \pi - \frac{\pi }{6}} \right)} \right)} $. For even $\lambda

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(C) Let $S = \sum\limits_{\lambda = 1}^{10} {{{\sin }^{ - 1}}\left( {\sin \left( {\lambda \pi - \frac{\pi }{6}} ight)} ight)} $. For even $\lambda = 2n$ (where $n = 1, 2, 3, 4, 5$): $\sin(2n\pi - \frac{\pi }{6}) = \sin(-\frac{\pi }{6}) = -\frac{1}{2}$. Thus,${{\sin }^{ - 1}}(\sin(2n\pi - \frac{\pi }{6})) = -\frac{\pi }{6}$. There are $5$ such terms,so the sum is $5 	imes (-\frac{\pi }{6}) = -\frac{5\pi }{6}$. For odd $\lambda = 2n-1$ (where $n = 1, 2, 3, 4, 5$): $\sin((2n-1)\pi - \frac{\pi }{6}) = \sin(\pi - \frac{\pi }{6}) = \sin(\frac{\pi }{6}) = \frac{1}{2}$. Thus,${{\sin }^{ - 1}}(\sin((2n-1)\pi - \frac{\pi }{6})) = \frac{\pi }{6}$. There are $5$ such terms,so the sum is $5 	imes \frac{\pi }{6} = \frac{5\pi }{6}$. Adding these together: $S = -\frac{5\pi }{6} + \frac{5\pi }{6} = 0$.

$\sum\limits_{\lambda = 1}^{10} {{{\sin }^{ - 1}}\left( {\sin \left( {\lambda \pi - \frac{\pi }{6}} \right)} \right)} $ is equal to-

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