`lim_(z->((2n+1)pi)/2)[z-(2n+1)pi/2]xx z/(cosz)`

Evaluate the integrals. int sqrt(1+sec x)dx on [(2n -(1)/(2))pi, (2n +(1)/(2)) pi], (n in Z).

The function f(x)=tan x is discontinuous on the set {n pi;n in Z}{(2n+1)(pi)/(2):n in Z}(d){(n pi)/(2):n in Z}

If n=5 in (i),then value of (z+1)(z^(2)-2z cos((pi)/(5))+1)(z^(2)+2z cos(2(pi)/(5))+1)

If A=[cos theta-sin theta sin theta cos theta], then A^(T)+A=I_(2), if theta=n pi,n in Z(b)theta=(2n+1)(pi)/(2),quad n in Z(c)theta=2n pi+(pi)/(3),quad n in Z(d) none of these

A general solution of tan^(2)theta+cos2 theta=1 is (n in Z)n pi=(pi)/(4) (b) 2n pi+(pi)/(4)n pi+(pi)/(4) (d) n pi

Let 1, z_(1),z_(2),z_(3),…., z_(n-1) be the nth roots of unity. Then prove that (1-z_(1))(1 - z_(2)) …. (1-z_(n-1))= n . Also,deduce that sin .(pi)/(n) sin.(2pi)/(pi)sin.(3pi)/(n)...sin.((n-1)pi)/(n) = (pi)/(2^(n-1))

Let 1, z_(1),z_(2),z_(3),…., z_(n-1) be the nth roots of unity. Then prove that (1-z_(1))(1 - z_(2)) …. (1-z_(n-1))= n . Also,deduce that sin .(pi)/(n) sin.(2pi)/(pi)sin.(3pi)/(n)...sin.((n-1)pi)/(n) = (pi)/(2^(n-1))

Let 1, z_(1),z_(2),z_(3),…., z_(n-1) be the nth roots of unity. Then prove that (1-z_(1))(1 - z_(2)) …. (1-z_(n-1))= n . Also,deduce that sin .(pi)/(n) sin.(2pi)/(pi)sin.(3pi)/(n)...sin.((n-1)pi)/(n) = (pi)/(2^(n-1))