For a positive integer n- let fn- tan -2 1-sec -1-sec 2-1-sec 4 -1-sec 2n - Then

(a,b,c,d) f_n(θ)=sin (θ/2)/cos ( θ/2 )[2 cos^2 θ/2/cos θ.2 cos^2 θ/cos 2θ.2 cos^2 θ/cos 4 θ.. ] Combine first two factors, f_n(θ)=sin θ/cos θ[2 cos^2 θ/cos 2 ...

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Byju's Answer

Standard XIII

Mathematics

Trigonometric Ratios of Multiples of an Angle

For a positiv...

Question

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

$f_{2} (\frac{π}{16}) = 1$

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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 option is D
$f_{5} (\frac{π}{128}) = 1$

$(a, b, c, d) f_{n} (θ) = \frac{s i n (\frac{θ}{2})}{c o s (\frac{θ}{2})} [\frac{2 c o s^{2} \frac{θ}{2}}{c o s θ} . \frac{2 c o s^{2} θ}{c o s 2 θ} . \frac{2 c o s^{2} θ}{c o s 4 θ} . .]$

Combine first two factors, $f_{n} (θ) = \frac{s i n θ}{c o s θ} [\frac{2 c o s^{2} θ}{c o s 2 θ} 3 \frac{2 c o s^{2} 2 θ}{c o s 4 θ} . . .]$

Again combine first two factors,

$f_{n} (θ) = t a n 2 θ [\frac{2 c o s^{2} 2 θ}{c o s 4 θ} . . .] = t a n (2^{n} θ) ∴ f_{2} (\frac{π}{16}) = t a n \frac{4 π}{16} = t a n (\frac{π}{4}) = 1 f_{3} (\frac{π}{32}) = t a n \frac{8 π}{32} = t a n (\frac{π}{4}) = 1$

$f_{4} (\frac{π}{64}) = t a n \frac{16 π}{64} = t a n (\frac{π}{4}) = 1 f_{5} (\frac{π}{128}) = t a n 32 \frac{π}{128} = t a n (\frac{π}{4}) = 1$

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