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For a positive integer n, fr(O) = () (1 ...

For a positive integer n, fr(O) = () (1 + sec ) (1 + sec 20)(1 + sec 40).... (1 + sec 2"0.) , then

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1 + sec20 =

(sec 2A+1 ) sec ^(2) A=

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)) .

For n in N, let f_(n) (x) = tan ""(x)/(2) (1+ sec x ) (1+ sec 2x) (1+ sec 4x)……(1+ sec 2 ^(n)x), the lim _(xto0) (f _(n)(x))/(2x) is equal to :

For n in N, let f_(n) (x) = tan ""(x)/(2) (1+ sec x ) (1+ sec 2x) (1+ sec 4x)……(1+ sec 2 ^(n)x), the lim _(xto0) (f _(n)(x))/(2x) is equal to :

For n in N, let f_(n) (x) = tan ""(x)/(2) (1+ sec x ) (1+ sec 2x) (1+ sec 4x)……(1+ sec 2 ^(n)x), the lim _(xto0) (f _(n)(x))/(2x) is equal to :

(1+ sec 20^(0))(1+sec40^(0))(1+ sec80^(0))=