[" 41.Let "],[qquad f_(n)(theta)=tan(theta)/(2)(1+sec theta)(1+sec2 theta)(1+sec4 theta)......(1+sec2^(n)theta)],[" Then "]

For a positive integer n, let : f_(n)(theta)=(tan.(theta)/(2))(1+sectheta)(1+sec2theta) (1+sec4theta)....(1+sec2^(n)theta) . Then :

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)) .

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

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

The value of tan((theta)/(2))(1+sec theta)(1+sec2 theta)+(1+sec2^(2)theta)...(1+sec2^(n)theta) is

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

For +ve integer n, let f_n(theta)=tan(theta/2)(1+sectheta)(1+sec2theta)(1+sec4theta)…..(1+sec2^ntheta) then

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) =

Prove that (tan 8 theta)/(tan theta)=(1+ sec 2 theta)(1+ sec 4 theta)(1+ sec 8 theta)