Q. if `f(n,theta)=prod_(r=1)^n(1-tan^2(theta/2^r))`, then compute `lim_(n->oo)f(n,theta)`

if f(n,theta)=prod_(r=1)^(n) (1-tan^(2)(theta)/(2^(r))) then the value of lim_(thetato0) [lim_(ntoinfty) f(n,theta)] , (where [x] represent the greatest integer function) is

If f(n)=prod_(r=1)^(n)cos r,n in N , then

If f (theta)=4/3 (1- cos ^(6) theta - sin ^(6)theta), then lim _(ntooo) 1/n [sqrt(f ((1)/(n)))+sqrt(f ((2)/(n)))+sqrt(f((n)/(n)))]=

If alpha, beta be the roots of ax^(2)+bx+c=0(a, b, c in R), (c )/(a)lt 1 and b^(2)-4ac lt 0, f(n)= sum_(r=1)^(n)|alpha|^(r )+|beta|^(r ) , then lim_(n to oo)f(n) is equal to

Let f(n)=prod_(r=1)^(n) sin(r), where r is in radians, then

If f (theta ) = lim _(x to oo) sum _( r =o) ^(n theta) (2r)/(n sqrt((3 thetan-2r)(n theta +2r))) then:

Let f(theta)=1/(tan^2theta){(1+tantheta)^3+(2+tantheta)^3+......+(10+tan theta)^3}-10 tan theta Then, lim_(theta->pi/2)f(theta) is equal to

sum_(n=1)^ootan(theta/(2^n))/(2^(n-1)cos(theta/(2^(n-1)))) is

For theta_(1), theta_(2), ...., theta_(n) in (0, pi/2) , if In (sec theta_(1)-tan theta_(1))+ "In" (sec theta_(2)-tan theta_(2))+...+"In" (sec theta_(n)-tan theta_(n))+ "ln" pi=0 , then the value of cos((sec theta_(1)+ tan theta_(1))(sec theta_(2)+tan theta_(2))....(sec theta_(n)+tan theta_(n))) is equal to

If f(x) = lim_(n->oo) sum_(r=0)^n (tan(x/2^(r+1)) + tan^3 (x/2^(r+1)))/(1- tan^2 (x/2^(r+1))) then lim_(x->0) f(x)/x is