[21quad " Single Answer Type "],[" Let "],[f_(n)(theta)=tan(theta)/(2)(1+sec theta)(1+sec2 theta)(1],[" then "],[" B Review Later "],[" d "f_(3)((pi)/(16))=1],[" D "f_(5)((pi)/(128))=1]

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)) .

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

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

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) =

(tan theta+sec theta-1)(tan theta+1+sec theta)=(2sin theta)/(1-sin 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

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))

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