For a positive integer n- let fn - - tan- 2 1 - - 1 - 2- 1 - 22- 1 - 2n- - then

The correct options are A f_2( π16) = 1 B f_3( π32) = 1 C f_4( π64) = 1 D f_5( π128) = 1 ...

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Byju's Answer

Standard XII

Mathematics

Sum of Cosines of Angles in Arithmetic Progression

For a positiv...

Question

For a positive integer n, let $f_{n} (θ) = (tan \frac{θ}{2}) (1 + sec θ) (1 + sec 2 θ) (1 + sec 2^{2} θ) . . . . . (1 + sec 2^{n} θ)$ , then

f_{2} (\frac{π}{16}) = 1

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f_{3} (\frac{π}{32}) = 1

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f_{4} (\frac{π}{64}) = 1

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f_{5} (\frac{π}{128}) = 1

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Solution

The correct options are
A $f_{2} (\frac{π}{16}) = 1$
B $f_{3} (\frac{π}{32}) = 1$
C $f_{4} (\frac{π}{64}) = 1$
D $f_{5} (\frac{π}{128}) = 1$

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For a positive integer n, let fn(θ)=(tanθ2)(1+secθ)(1+sec2θ)(1+sec22θ).....(1+sec2nθ) , then

The correct options are A f2(π16)=1 B f3(π32)=1 C f4(π64)=1 D f5(π128)=1

For a positive integer n, let $f_{n} (θ) = (tan \frac{θ}{2}) (1 + sec θ) (1 + sec 2 θ) (1 + sec 2^{2} θ) . . . . . (1 + sec 2^{n} θ)$ , then

The correct options are
A $f_{2} (\frac{π}{16}) = 1$
B $f_{3} (\frac{π}{32}) = 1$
C $f_{4} (\frac{π}{64}) = 1$
D $f_{5} (\frac{π}{128}) = 1$