For a positive integer n- let f n -tan-21-1- 2 -1- 4 - -1- 2 n -ThenA- f 2-16-1B- f3-32-1C- f4-64-1D- f 5-128-1

The correct options are A f_2 ( π/16 )=1 B f_3 ( π/32 )=1 C f_4 ( π/64 )=1 D f_5 ( π/128 )=1 E=f_n(θ)=sin ( θ/2 )/cos ( θ/2 ) [ 2cos^2 ( θ/2 )/cosθ .2cos^2 ...

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Byju's Answer

Standard XI

Mathematics

Trigonometric Ratios of Multiples of an Angle

For a positiv...

Question

For a positive integer n, let
$f_{n} (θ) = (t a n \frac{θ}{2}) (1 + s e c θ) (1 + s e c 2 θ) (1 + s e c 4 θ) \dots (1 + s e c 2^{n} θ)$ .
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
$f_{5} (\frac{π}{128}) = 1$

$E = f_{n} (θ) = \frac{s i n (\frac{θ}{2})}{c o s (\frac{θ}{2})} [\frac{2 c o s^{2} (\frac{θ}{2})}{c o s θ} . \frac{2 c o s^{2} θ}{c o s 2 θ} . \frac{2 c o s^{2} 2 θ}{c o s 4 θ} \dots \frac{2 c o s^{2} 2^{n - 1} θ}{c o s 2^{n} θ}] = \frac{s i n θ}{c o s θ} [\frac{2 c o s^{2} θ}{c o s 2 θ} . \frac{2 c o s^{2} 2 θ}{c o s 4 θ} \dots \frac{2 c o s^{2} 2^{n - 1} θ}{c o s 2^{n} θ}] = t a n 2^{n} θ . n = 2, θ = \frac{π}{16} f_{2} (\frac{π}{16}) = t a n 4. \frac{π}{16} = t a n \frac{π}{4} = 1$
Similarly, $f_{3} (\frac{π}{32}), f_{4} (\frac{π}{64}) a n d f_{5} (\frac{π}{128}) i s t a n \frac{π}{4} = 1$

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For a positive integer n, let fn(θ)=(tanθ2)(1+secθ)(1+sec2θ)(1+sec4θ)…(1+sec2nθ). Then

For a positive integer n, let
$f_{n} (θ) = (t a n \frac{θ}{2}) (1 + s e c θ) (1 + s e c 2 θ) (1 + s e c 4 θ) \dots (1 + s e c 2^{n} θ)$ .
Then