The number of positive integers which satisfy `sin(pi/(4n))+cos (pi/(4n))=sqrt(n/2` is -

Let n be a positive integer such that sin(pi/(2n))+cos(pi/(2n))=(sqrt(n))/2 Then

Let n be a fixed positive integer such that sin(pi/(2n))+cos(pi/(2n))=sqrtn/2 then

Let n be a positive integer such that sin(pi/(2n))+cos(pi/(2n))=(sqrt(n))/2dot Then

Let n be a positive integer such that sin (pi/(2n))+cos(pi/(2n))=(sqrt(n))/2dot Then

Let n be a positive integer such that sin (pi/(2n))+cos (pi/(2n))= sqrt(n)/2 then (A) n=6 (B) n=1,2,3,….8 (C) n=5 (D) none of these

Let n be a positive integer such that sin (pi/(2n))+cos (pi/(2n))= sqrt(n)/2 then (A) n=6 (B) n=1,2,3,….8 (C) n=5 (D) none of these

Let n be a positive integer such that sin((pi)/(2^(n)))+cos((pi)/(2^(n)))=(sqrt(n))/(2), then

Let n be a positive integer such that (sin pi)/(2n)+(cos pi)/(2n)=(sqrt(n))/(2)* Then 6<=n<=8(b)4

If n is positive integer and "cos" (pi)/(2n)+"sin" (pi)/(2n) =(sqrt(n))/(2) , then prove that 4 le n le 8

The least positive integer n such that [{:(cos""(pi)/(4), sin""(pi)/(4)) ,(- sin ""(pi)/(4) , cos ""(pi)/(4)):}]^(n) is an identity matrix of order 2 is