`f_n(theta)=sum_(n=1)^n tan(theta/2^n)sec(theta/(2^(n-1))),`then compute the value of `f_4( (4pi)/3),`

f_(n)(theta)=sum_(n=1)^(n)tan((theta)/(2^(n)))sec((theta)/(2^(n-1))), then compute the value of f_(4)((4 pi)/(3))

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

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

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

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(theta)=sum_(n=1)^(6)cos ec(theta+((n-1)pi)/(4))cos ec(theta+(npi)/(4)) , where 0ltthetalt(pi)/(2) , then find the minimum value of f(theta) .

For positive integer n if f_(n)(theta) = tan ""(theta)/(2) (1 + sec theta) (1+ sec2 theta) (1 + sec4 theta)"…."( 1 + sec2^(n)theta) then find the value of f_(2) ((pi)/(16)) " and " f_(5) ((pi)/(128)) .