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`

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

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)

sec(pi/4+theta)sec(pi/4-theta)=2sec2theta

Find the value of tantheta(1+sec2theta)(1+sec4theta)(1+sec8theta) , when theta=pi/32

Find the value of tantheta(1+sec2theta)(1+sec4theta)(1+sec8theta) , when theta=pi/32

Prove that: (1+sec2theta)(1+sec4theta)(1+sec8theta)...(1+sec2^ntheta)=t a n2^nthetacottheta, n in Ndot

Prove that: (1+sec2theta)(1+sec4theta)(1+sec8theta)....(1+sec2^ntheta)=t a n2^nthetacottheta, n in Ndot

Let f_n(theta)=(cos(theta/2)+cos2theta+cos((7theta)/2)+....+cos(3n-2)(theta/2))/(sin(theta/2)+sin2theta+sin((7theta)/2)+....+sin(3n-2)(theta/2)) then (a) f_3((3pi)/(16))=sqrt(2)-1 (b) f_5(pi/(28))=sqrt(2)+1 (c) f_7(pi/(60))=(2+sqrt(3)) (d) none of these

Let f(theta)=1/(tan^2theta){(1+tantheta)^3+(2+tantheta)^3+......+(10+tan theta)^3}-10 tan theta Then, lim_(theta->pi/2)f(theta) is equal to

If theta=pi/(4n) then the value of tanthetatan(2theta)tan(3theta)....tan((2n-1)theta) is