[f_(n)(theta)=tan((theta)/(2))*(1+sec theta)(1+sec2 theta)...(1+sec2'2 theta)],[" find "f_(5)((pi)/(28))f_(5)((pi)/(128))]

f_(n)(Theta)=(tan((Theta)/(2)))(1+sec(Theta))(1+sec(2Theta))......(1+sec(2^(n)Theta)

1+(tan^(2)theta)/((1+sec theta))=sec theta

Let f_(n)(theta)= tan""(theta)/2(1+sectheta)(1+sec2theta)(1+sec4theta).....(1+sec2^(n) theta) ,then f_(2)(pi/16)+f_(3)((pi)/32)+f_(4)(pi/64)+f_5(pi/128) =

f(n,theta)=(1+sec2 theta)(1+sec2^(2)theta)...(1+sec2^(n)theta) then f(1,(pi)/(6)) and f(2,(pi)/(12))=

Let f_n(Theta)=(tan(Theta/2))(1 +sec(Theta))(1+sec(2Theta)).....(1+sec(2^nTheta) .. then

Prove that (tan8 theta)/(tan theta)=(1+sec2 theta)(1+sec4 theta)(1+sec8 theta)

(tan theta)/(sec theta+1)+(sec theta+1)/(tan theta)=2cos ec theta

The value of tan(theta/2)(1+sectheta)(1+sec2theta)+(1+sec2^2theta)....(1+sec2^ntheta) is

(tan theta)/(sec theta-1)+(tan theta)/(sec theta+1)=2cos ec| theta

For a positive integer n f_(n)(theta)=((tan theta)/(2))(1+sec theta)(1+sec2 theta)(1+sec4 theta)...(1+sec2^(n)theta) ,then