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$I f n \in N, \sum_{k = 1}^{n} \sin^{- 1} (x_{_{K}}) = \frac{n π}{2}$ then $\sum_{k = 1}^{n} x_{_{K}} =$

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Correct option is A

∑k=1nsin-1xk=nπ2⇒sin-1xk=π2 for k=1,2,...n⇒xk=sinπ2=1 for k=1,2,...n∴∑k=1nxk=1+1+1+......+1n times=n

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If n∈N, ∑k=1nsin−1x K=nπ2then∑k=1nx K=

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$I f n \in N, \sum_{k = 1}^{n} \sin^{- 1} (x_{_{K}}) = \frac{n π}{2}$ then $\sum_{k = 1}^{n} x_{_{K}} =$