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sum(i=1)^(2n) sin^(-1)(xi)=npi then the ...

`sum_(i=1)^(2n) sin^(-1)(x_i)=npi` then the value of `sum_(i=1)^n cos^(-1)x_i+sum_(i=1)^n tan^(-1)x_i=` (A) `(npi)/4` (B) `(2/3)npi` (C) `(5/4)npi` (D) `2npi`

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