Let fn (θ)=tanθ2(1+secθ)(1+sec2θ)(1+sec4θ)................(1+sec2nθ), then f2π16+f3π32+f4π64+f5π128 =

fn(θ)=tan θ2 (1+secθ) (1+sec2θ) =sin θ2cos θ2·1+cosθcosθ·1+cos2θcos2θ......(1+cos2nθ)cos2nθ =sin θ2cos θ2··2cos2 θ2cosθ·2cos2θcos2θ.....2cos2n-1θcos2nθ =tan2nθf2 π16=tan 4π16=tanπ4=1 f3 π32=tan 8π32=tanπ4=1 f4 π64=tan 16π64=tanπ4=1 f5 π128=tan 32π128=tanπ4=1 f2 π16+f3 π32+f4 π64+f5 π128=4

Let fn (θ)=tanθ2(1+secθ)(1+sec2θ)(1+sec4θ)................(1+sec2nθ), then f2π16+f3π32+f4π64+f5π128 =

fn(θ)=tan θ2 (1+secθ) (1+sec2θ) =sin θ2cos θ2·1+cosθcosθ·1+cos2θcos2θ......(1+cos2nθ)cos2nθ =sin θ2cos θ2··2cos2 θ2cosθ·2cos2θcos2θ.....2cos2n-1θcos2nθ =tan2nθf2 π16=tan 4π16=tanπ4=1 f3 π32=tan 8π32=tanπ4=1 f4 π64=tan 16π64=tanπ4=1 f5 π128=tan 32π128=tanπ4=1 f2 π16+f3 π32+f4 π64+f5 π128=4

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$L e t f_{n} (θ) = \tan \frac{θ}{2} (1 + s e c θ) (1 + s e c 2 θ) (1 + s e c 4 θ) . . . . . . . . . . . . . . . . (1 + s e c 2^{n} θ), t h e n f_{2} (\frac{π}{16}) + f_{3} (\frac{π}{32}) + f_{4} (\frac{π}{64}) + f_{5} (\frac{π}{128}) =$

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Detailed Solution

$\begin{array}{rcl} f_{n} (θ) & = & \tan \frac{θ}{2} (1 + secθ) (1 + \sec 2 θ) \\ = & \frac{\sin \frac{θ}{2}}{\cos \frac{θ}{2}} \cdot \frac{1 + cosθ}{cosθ} \cdot \frac{1 + \cos 2 θ}{\cos 2 θ} . . . . . . \frac{(1 + \cos 2^{n} θ)}{\cos 2^{n} θ} \\ = & \frac{\sin \frac{θ}{2}}{\cos \frac{θ}{2}} \cdot \cdot \frac{2 \cos^{2} \frac{θ}{2}}{cosθ} \cdot \frac{2 \cos^{2} θ}{\cos 2 θ} . . . . . \frac{2 \cos 2^{n - 1} θ}{\cos 2^{n} θ} \\ = & \tan 2^{n} θ \end{array}$

$f_{2} (\frac{π}{16}) = \tan (\frac{4 π}{16}) = \tan \frac{π}{4} = 1 f_{3} (\frac{π}{32}) = \tan (\frac{8 π}{32}) = \tan \frac{π}{4} = 1 f_{4} (\frac{π}{64}) = \tan (\frac{16 π}{64}) = \tan \frac{π}{4} = 1 f_{5} (\frac{π}{128}) = \tan (\frac{32 π}{128}) = \tan \frac{π}{4} = 1 f_{2} (\frac{π}{16}) + f_{3} (\frac{π}{32}) + f_{4} (\frac{π}{64}) + f_{5} (\frac{π}{128}) = 4$