Let fn - -tan-2- 1-sec - 1-sec 2- 1-sec 4- - 1-sec 2- - Then

The correct options are A f_2 ( π/16 )=1 B f_3 ( π/32 )=1 C f_5 ( π/128 )=1 D f_4 ( π/64 )=1 nbsp; f _ n ( θ nbsp;) =sin( θ nbsp ...

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Standard X

Mathematics

Standard Values of Trigonometric Ratios

Let fn θ =t...

Question

Let $f_{n} (θ) = t a n \frac{θ}{2} . (1 + s e c θ) (1 + s e c 2 θ) (1 + s e c 4 θ) . . . (1 + s e c 2^{''} θ) .$ 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 $f_{5} (\frac{π}{128}) = 1$
D $f_{4} (\frac{π}{64}) = 1$
$f_{n} (θ) = \frac{sin (\frac{θ}{2})}{cos (\frac{θ}{2})} \times \frac{1 + cos θ}{cos θ} (1 + sec 2 θ) . . . (1 + sec 2^{n} θ)$
$= \frac{sin (\frac{θ}{2}) \times {cos}^{2} (\frac{θ}{2})}{cos (\frac{θ}{2}) cos θ} (1 + sec 2 θ) . . . (1 + sec 2^{n} θ)$
$= tan θ (1 + sec 2 θ) (1 + sec 4 θ) . . . (1 + sec 2^{n} θ) = tan 2^{n} θ$
Therefore
$f_{2} (\frac{π}{16}) = f_{3} (\frac{π}{32}) = f_{4} (\frac{π}{64}) = f_{5} (\frac{π}{128}) = tan \frac{π}{4}$

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Similar questions

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

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

Q. 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) .

Then

Q. For a positive integer

n,

let

f_{n} (θ) = (tan \frac{θ}{2}) (1 + sec θ) (1 + sec 2 θ) (1 + sec 4 θ) . . . (1 + sec 2^{n} θ) .

Then

Q. If

f_{n} (θ) = n \sum r = 0 \frac{1}{4^{r}} {sin}^{4} (2^{r} θ)

, then which of the following alternative(s) is/are correct?

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Let fn(θ)=tanθ2.(1+secθ)(1+sec2θ)(1+sec4θ)...(1+sec2′′θ). Then

The correct options are A f2(π16)=1 B f3(π32)=1 C f5(π128)=1 D f4(π64)=1fn(θ)=sin(θ2)cos(θ2)×1+cosθcosθ(1+sec2θ)...(1+sec2nθ)=sin(θ2)×cos2(θ2)cos(θ2)cosθ(1+sec2θ)...(1+sec2nθ)=tanθ(1+sec2θ)(1+sec4θ)...(1+sec2nθ)=tan2nθThereforef2(π16)=f3(π32)=f4(π64)=f5(π128)=tanπ4

Let $f_{n} (θ) = t a n \frac{θ}{2} . (1 + s e c θ) (1 + s e c 2 θ) (1 + s e c 4 θ) . . . (1 + s e c 2^{''} θ) .$ Then