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

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

prove that : (tan 2^n theta)/(tan theta) = (1+sec 2theta) (1+ sec 2^2 theta) (1+sec 2^3 theta) … (1+ sec 2^n theta)

prove that : (tan 2^n theta)/(tan theta) = (1+sec 2theta) (1+ sec 2^2 theta) (1+sec 2^3 theta) … (1+ sec 2^n theta)

(1 + sec 2 theta)(1 + sec 2^(2) theta)(1 + sec 2^(3) is equal is

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)) .

(sec^2 theta-1)/(sec^2 theta)- sin^2 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) =

For a positive integer n , let f_n(theta)=(tantheta//2)(1+sectheta)(1+sec2theta)(1+sec4theta).......... (1+sec2^ntheta) . Then (a) f_2(pi/(16))=1 (b) f_3(pi/(32))=1 (c) f_4(pi/(64))=1 (d) f_5(pi/(128))=1

For a positive integer n , let f_n(theta)=(tantheta//2)(1+sectheta)(1+sec2theta)(1+sec4theta).......... (1+sec2^ntheta) . Then (a) f_2(pi/(16))=1 (b) f_3(pi/(32))=1 (c) f_4(pi/(64))=1 (d) f_5(pi/(128))=1

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