Let f_n (θ) = \tan \theta/2 (1 + sec θ) (1 + sec 2θ) (1 + sec 4θ) ... (+\sec 2^n \theta) then f_2(\pi/{16}) + f_3(\pi/{32}) + f_4(\pi/{64}) + f_5(\pi/{128}) is

JEE Mains 2020 Model Paper 4 New Pattern

Question : 54 of 75

Marks: +1, -0

Let fn (θ) = tan

(1 + sec θ) (1 + sec 2θ) (1 + sec 4θ) ... (+sec2nθ) then
f2(

)+f3(

)+f4(

)+f5(

128

) is

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