If `pi/2

Find the value of expression ((cos pi)/(2)+i sin((pi)/(2)))(cos((pi)/(2^(2)))+i sin((pi)/(2^(2))))......oo

The largest interval lying in (-pi/2,pi/2) for which the function [f(x)=4^-x^2+cos^(-1)(x/2-1)+log(cosx)] is defined, is (1) [0,pi] (2) (-pi/2,pi/2) (3) [-pi/4,pi/2) (4) [0,pi/2)

If y=tan^(-1)x+tan^(-1)(1/x)+sec^(-1)x , then y lies in the interval (a) [pi/2,pi)uu[pi,(3pi)/2) (b) [pi/2,(3pi)/2] (c)(0,pi) (d) [0,pi/2)uu[pi/2,pi)

f(x)={((5^(cosx)-1)/((pi)/(2)-x)",", x ne (pi)/(2)), (log 5"," , x =(pi)/(2)):} at x =(pi)/(2) is

If f(x) = {((1-sinzx)/(pi-2x), ",", x != pi/2),(lambda, ",", x = pi/2):} , be continuous at x = (pi)/2 , then the value of lambda is

If f(x) is continuous at x = (pi)/2 , where f(x) = (sinx)^(1/(pi-2x)), "for" x != (pi)/2 , then f((pi)/(2)) =

sin((pi)/(2)+theta)+sin((3 pi)/(2)+theta)+sin((5 pi)/(2)+theta)+sin((7 pi)/(2)+theta) is equal to

f(x) = {{:((k cosx )/((pi - 2x)"," if x ne (pi)/(2))),(3"," if x = (pi)/(2)):} at x = (pi)/(2) .

If : f(x) {: ((1-sin x)/(pi-2x)", ... " x != pi//2 ),(=lambda ", ... " x = pi//2):} is continuous at x = pi//2 , then : lambda =

The argument of (1-i)/(1+i) is (pi)/(2) b.(pi)/(2) c.(3 pi)/(2)d .(5 pi)/(2)